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#### Question 1:

For the following arithmetic progressions write the first term a and the common difference d:

(i) −5, −1, 3, 7, ...

(ii) $\frac{1}{5},\frac{3}{5},\frac{5}{5},\frac{7}{5}$

(iii) 0.3, 0.55, 0.80, 1.05, ...

(iv) −1.1, −3.1, −5.1, −7.1, ...

In the given problem, we need to write the first term (a) and the common difference (d) of the given A.P

(i) −5, −1, 3, 7 …

Here, first term of the given A.P is (a) = −5

Now, we will find the difference between the two terms of the given A.P

Similarly,

Also,

As

Therefore, the first term of the given A.P is and the common difference of the given is

(ii)

Here, first term of the given A.P is (a) =

Now, we will find the difference between the two terms of the given A.P

Similarly,

Also,

As

Therefore, the first term of the given A.P is and the common difference is

(iii) 0.3, 0.55, 0.80, 1.05, …

Here, first term of the given A.P is (a) = 0.3

Now, we will find the difference between the two terms of the given A.P

Similarly,

Also,

As

Therefore, the first term of A.P is and the common difference is

(iv) −1.1, −3.1, −5.1, −7.1...

Here, first term of the given A.P is (a) = −1.1

Now, we will find the difference between the two terms of the given A.P

Similarly,

Also,

As

Therefore, the first term of A.P is and the common difference is

#### Question 2:

Write the arithmetic progression when first term a and common difference d are as follows:

(i) a = 4, d = −3

(ii) a = −1, d = $\frac{1}{2}$

(iii) a = −1.5, d = −0.5

In the given problem, we are given its first term (a) and common difference (d).

We need to find the A.P

(i) ,

Now, as

A.P would be represented by

So,

Similarly,

Also,

Further,

Therefore, A.P with and is

(ii) ,

Now, as

A.P would be represented by

So,

Similarly,

Also,

Further,

Therefore, A.P with and is

(iii) ,

Now, as

A.P would be represented by

So,

Similarly,

Also,

Further,

Therefore, A.P with and is

#### Question 3:

In which of the following situations, the sequence of numbers formed will form an A.P.?

(i) The cost of digging a well for the first metre is Rs 150 and rises by Rs 20 for each succeeding metre.

(ii) The amount of air present in the cylinder when a vacuum pump removes each time $\frac{1}{4}$ of their remaining in the cylinder.

(iii) Divya deposited Rs 1000 at compound interest at the rate of 10% per annum. The amount at the end of first year, second year, third year, ... and so on.

(i) In the given problem,

Cost of digging a well for the first meter = Rs 150

Cost of digging a well for subsequent meter is increased by Rs 20

So,
Cost of digging a well of depth one meter= Rs. 150

Cost of digging a well of depth two meters= Rs = Rs.

Cost of digging a well of depth three meters= Rs = Rs.

Cost of digging a well of depth four meters = Rs = Rs.

Thus, the costs of digging a well of different depths are

Now, for a sequence to be an A.P., the difference between adjacent terms should be equal.

Here,

Also,

Therefore,

Since the terms of the sequence are at a common difference of 20, the above sequence is an A.P. with the first term asand common difference.

(ii) Here, let us take the initial amount of air present in the cylinder as 100 units.

So,

Amount left after vacuum pump removes air for 1st time=

Amount left after vacuum pump removes air for 2nd time=

Amount left after vacuum pump removes air for 3rd time=

Thus, the amount left in the cylinder at various stages is

Now, for a sequence to be an A.P., the difference between adjacent terms should be equal.

Here,

Also,

Since,

The sequence is not an A.P.

(iii)
Here, prinical (P) = 1000
Rate (r) = 10%
Amount compounded annually is given by
$A=P{\left(1+\frac{r}{100}\right)}^{n}$
For the first year,
${A}_{1}=1000{\left(1+\frac{10}{100}\right)}^{1}=1100$
For the second year,
${A}_{2}=1000{\left(1+\frac{10}{100}\right)}^{2}=1210$
For the third year,
${A}_{1}=1000{\left(1+\frac{10}{100}\right)}^{3}=1331$
Therefore, first three terms are 1100, 1210, 1331.
The common difference between the consecutive terms are not same.
Hence, this is not in A.P.

#### Question 4:

Find the common difference and write the next four terms of each of the following arithmetic progressions:

(i) 1, −2, −5, −8, ...

(ii) 0, −3, −6, −9, ...

(iii)

(iv)

In the given problem, we need to find the common difference and the next four terms of the given A.P.

(i)

Here, first term (a1) =1

Common difference (d)

Now, we need to find the next four terms of the given A.P

That is we need to find

So, using the formula

Substituting in the above formula

Substituting n = 5, we get

Substituting n = 6, we get

Substituting n = 7, we get

Substituting n = 8, we get

Therefore, the common difference is and the next four terms are

(ii)

Here, first term (a1) =0

Common difference (d)

Now, we need to find the next four terms of the given A.P

That is we need to find

So, using the formula

Substituting in the above formula

Substituting n = 5, we get

Substituting n = 6, we get

Substituting n = 7, we get

Substituting n = 8, we get

Therefore, the common difference is and the next four terms are

(iii)

Here, first term (a1) =−1

Common difference (d)

Now, we need to find the next four terms of the given A.P

That is we need to find

So, using the formula

Substituting in the above formula

Substituting n = 4, we get

Substituting n = 5, we get

Substituting n = 6, we get

Substituting n = 7, we get

Therefore, the common difference is and the next four terms are

(iv)

Here, first term (a1) =−1

Common difference (d)

Now, we need to find the next four terms of the given A.P

That is we need to find

So, using the formula

Substituting in the above formula

Substituting n = 4, we get

${a}_{4}=-1+\left(4-1\right)\left(\frac{1}{6}\right)\phantom{\rule{0ex}{0ex}}{a}_{4}=-1+\left(\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}{a}_{4}=\frac{-2+1}{2}=\frac{-1}{2}$

Substituting n = 5, we get

Substituting n = 6, we get

Substituting n = 7, we get

Therefore, the common difference is and the next four terms are

#### Question 5:

Which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.

(i) 3, 6, 12, 24, ...

(ii) 0, −4, −8, −12, ...

(iii)

(iv) 12, 2, −8, −18, ...

(v) 3, 3, 3, 3, ...

(vi) p, p + 90, p + 180 p + 270, ... where p = (999)999

(vii) 1.0, 1.7, 2.4, 3.1, ...

(viii) −225, −425, −625, −825, ...

(ix) 10, 10 + 25, 10 + 26, 10 + 27,...

(x) a + b, (a + 1) + b, (a + 1) + (b + 1), (a + 2) + (b + 1), (a + 2) + (b + 2), ...

(xi) 12, 32, 52, 72, ...

(xii) 12, 52, 72, 73, ...

In the given problem, we are given various sequences.

We need to find out that the given sequences are an A.P or not and then find its common difference (d)

(i)

Here,

First term (a) = 3

Now, for the given to sequence to be an A.P,

Common difference (d

Here,

Also,

Since

Hence, the given sequence is not an A.P

(ii)

Here,

First term (a) = 0

Now, for the given sequence to be an A.P,

Common difference (d

Here

Also,

Since

Hence, the given sequence is an A.P and its common difference is

(iii)

Here,

First term (a) =

Now, for the given to sequence to be an A.P,

Common difference (d

Here,

Also,

Since

Hence, the given sequence is not an A.P

(iv)

Here,

First term (a) = 12

Now, for the given to sequence to be an A.P,

Common difference (d

Here,

Also,

Since

Hence, the given sequence is an A.P with the common difference

(v)

Here,

First term (a) = 3

Now, for the given to sequence to be an A.P,

Common difference (d

Here,

Also,

Since

Hence, the given sequence is an A.P and its common difference is

(vi) Where,

Here,

First term (a) = p

Now, for the given to sequence to be an A.P,

Common difference (d

Here,

Also,

Since

Hence, the given sequence is an A.P and its common difference is

(vii)

Here,

First term (a) = 1.0

Now, for the given to sequence to be an A.P,

Common difference (d

Here,

Also,

Since

Hence, the given sequence is an A.P and its common difference is

(viii)

Here,

First term (a) = −225

Now, for the given to sequence to be an A.P,

Common difference (d

Here,

Also,

Since

Hence, the given sequence is an A.P and its common difference is

(ix)

Here,

First term (a) = 10

Now, for the given to sequence to be an A.P,

Common difference (d) = ${a}_{1}-a={a}_{2}-{a}_{1}$

Here,

Also,

Since

Hence, the given sequence is not an A.P

(x)

Here,

First term (a) = + b

Now, for the given to sequence to be an A.P,

Common difference (d

Here,

Also,

Since

Hence, the given sequence is an A.P and its common difference is

(xi)

Here,

First term (a) =

Now, for the given to sequence to be an A.P,

Common difference (d

Here,

Also,

Since

Hence, the given sequence is not an A.P

(xii)

Here,

First term (a) =

Now, for the given to sequence to be an A.P,

Common difference (d

Here,

Also,

Since

Hence, the given sequence is an A.P with the common difference .

#### Question 6:

Find the common difference of the A.P. and write the next two terms:

(i) 51, 59, 67, 75, ..

(ii) 75, 67, 59, 51, ...

(iii) 1.8, 2.0, 2.2, 2.4, ...

(iv)

(v) 119, 136, 153, 170, ...

In this problem, we are given different A.P. and we need to find the common difference of the A.P., along with the next two terms.

(i)

Here,

So, common difference of the A.P. (d) =

Also, we need to find the next two terms of A.P., which means we have to find the 5th and 6th term.

So, for fifth term,

Similarly, we find the sixth term,

Therefore, the common difference is and the next two terms of the A.P. are.

(ii)

Here,

So, common difference of the A.P. (d) =

Also, we need to find the next two terms of A.P., which means we have to find the 5th and 6th term.

So, for fifth term,

Similarly, we find the sixth term,

Therefore, the common difference is and the next two terms of the A.P. are.

(iii)

Here,

So, common difference of the A.P. (d) =

Also, we need to find the next two terms of A.P., which means we have to find the 5th and 6th term.

So, for fifth term,

Similarly, we find the sixth term,

Therefore, the common difference is and the next two terms of the A.P. are.

(iv)

Here,

So, common difference of the A.P. (d) =

Also, we need to find the next two terms of A.P., which means we have to find the 5th and 6th term.

So, for fifth term,

Similarly, we find the sixth term,

Therefore, the common difference is and the next two terms of the A.P. are.

(v)

Here,

So, common difference of the A.P. (d) =

Also, we need to find the next two terms of A.P., which means we have to find the 5th and 6th term.

So, for fifth term,

Similarly, we find the sixth term,

Therefore, the common difference is and the next two terms of the A.P. are.

#### Question 7:

Prove that no matter what the real numbers a and b are, the sequence with nth term a + nb is always an A.P. What is the common difference?

In the given problem, we are given the sequence with the nth term () as where a and b are real numbers.

We need to show that this sequence is an A.P and then find its common difference (d)

Here,

Now, to show that it is an A.P, we will find its few terms by substituting

So,

Substituting n = 1, we get

Substituting n = 2, we get

Substituting n = 3, we get

Further, for the given to sequence to be an A.P,

Common difference (d)

Here,

Also,

Since

Hence, the given sequence is an A.P and its common difference is .

#### Question 1:

Find:

(i) 10th term of the A.P. 1, 4, 7, 10, ...

(ii) 18th term of the A.P.

(iii) nth term of the A.P. 13, 8, 3, −2, ...

(iv) 10th term of the A.P. −40, −15, 10, 35, ...

(v) 8th term of the A.P. 117, 104, 91, 78, ...

(vi) 11th term of the A.P. 10.0, 10.5, 11.0, 11.5, ...

(vii) 9th term of the A.P.

In this problem, we are given different A.P. and we need to find the required term of that A.P.

(i) 10th term of the A.P.

Here,

First term (a) = 1

Common difference of the A.P. (d)

Now, as we know,

So, for 10th term,

Therefore, the 10th term of the given A.P. is.

(ii) 18th term of the A.P.

Here,

First term (a) =

Common difference of the A.P. (d)

Now, as we know,

So, for 18th term,

Therefore, the 18th term of the given A.P. is.

(iii) nth term of the A.P.

Here,

First term (a) = 13

Common difference of the A.P. (d)

Now, as we know,

So, for nth term,

Therefore, the nth term of the given A.P. is.

(iv) 10th term of the A.P.

Here,

First term (a) = -40

Common difference of the A.P. (d)

Now, as we know,

So, for 10th term,

Therefore, the 10th term of the given A.P. is.

(v) 8th term of the A.P.

Here,

First term (a) = 117

Common difference of the A.P. (d)

Now, as we know,

So, for 8th term,

Therefore, the 8th term of the given A.P. is.

(vi) 11th term of the A.P.

Here,

First term (a) = 10.0

Common difference of the A.P. (d)

Now, as we know,

So, for 11th term,

Therefore, the 11th term of the given A.P. is.

(vii) 9th term of the A.P.

Here,

First term (a) =

Common difference of the A.P. (d)

Now, as we know,

So, for 9th term,

Therefore, the 9th term of the given A.P. is.

#### Question 2:

Find:

(i) Which term of the A.P. 3, 8, 13, ... is 248?

(ii) Which term of the A.P. 84, 80, 76, ... is 248?

(iii) Which term of the A.P. 4, 9, 14, ... is 254?

(iv) Which term of the A.P. 21, 42, 63, 84, ... is 420?

(v) Which term of the A.P. 121, 117, 113, ... is its first negative term?

(vi) Which term of the A.P. –7, –12, –17, –22,... will be –82? Is –100 any term of the A.P?

In the given problem, we are given an A.P and the value of one of its term.

We need to find which term it is (n)

So here we will find the value of n using the formula,

(i) Here, A.P is

Now,

Common difference (d) =

=

= 5

Thus, using the above mentioned formula

Thus,

Therefore 248 is the  of the given A.P

(ii) Here, A.P is

Now,

Common difference (d) =

=80-84

= -4

Thus, using the above mentioned formula

On further simplifying, we get,

Thus,

Therefore 84 is the  of the given A.P

(iii) Here, A.P is

Now,

Common difference (d) =

= 9-4

= 5

Thus, using the above mentioned formula

Thus,

Therefore 254 is the  of the given A.P

(iv) Here, A.P is

Now,

Common difference (d) =

= 42-21

= 21

Thus, using the above mentioned formula

Thus,

Therefore 420 is the  of the given A.P

(v) Here, A.P is

We need to find first negative term of the A.P

Now,

Common difference (d) =

Now, we need to find the first negative term,

Further simplifying, we get,

Thus,

Therefore, the first negative term is the of the given A.P.

(vi)  –7, –12, –17, –22,...

#### Question 3:

Find:

(i) Is 68 a term of the A.P. 7, 10, 13, ...?

(ii) Is 302 a term of the A.P. 3, 8, 13, ...?

(iii) Is − 150 a term of the A.P. 11, 8, 5, 2, ...?

In the given problem, we are given an A.P and the value of one of its term.

We need to find whether it is a term of the A.P or not.

So here we will use the formula,

(i) Here, A.P is

Now,

Common difference (d) =

Thus, using the above mentioned formula, we get,

Since, the value of n is a fraction.

Thus, 68 is not the term of the given A.P

(ii) Here, A.P is

Now,

Common difference (d) =

Thus, using the above mentioned formula, we get,

Since, the value of n is a fraction.

Thus, 302 is not the term of the given A.P

(iii) Here, A.P is

Now,

Common difference (d) =

Thus, using the above mentioned formula

Since, the value of n is a fraction.

Thus, -150 is not the term of the given A.P

#### Question 4:

How many terms are there in the A.P.?

(i) 7, 10, 13, ... 43.

(ii) $-1,-\frac{5}{6},-\frac{2}{3},-\frac{1}{2},...\frac{10}{3}.$

(iii) 7, 13, 19, ..., 205.

(iv)

In the given problem, we are given an A.P.

We need to find the number of terms present in it

So here we will find the value of n using the formula,

(i) Here, A.P is

The first term (a) = 7

The last term () = 43

Now,

Common difference (d) =

Thus, using the above mentioned formula, we get,

Thus,

Therefore, the number of terms present in the given A.P is .

(ii) Here, A.P is

The first term (a) = -1

The last term () =

Now,

Common difference (d) =

Thus, using the above mentioned formula, we get,

Further solving for n, we get

Thus,

Therefore, the number of terms present in the given A.P is .

(iii) Here, A.P is

The first term (a) = 7

The last term () = 205

Now,

Common difference (d) =

Thus, using the above mentioned formula, we get,

Thus,

Therefore, the number of terms present in the given A.P is

(iv) Here, A.P is

The first term (a) = 18

The last term () = -47

Now,

Common difference (d) =

Thus, using the above mentioned formula, we get,

Further, solving for n, we get

Thus,

Therefore, the number of terms present in the given A.P is .

#### Question 5:

The first term of an A.P. is 5, the common difference is 3 and the last term is 80;

In the given problem, we are given an A.P whose,

First term (a) = 5

Last term () = 80

Common difference (d) = 3

We need to find the number of terms present in it (n)

So here we will find the value of n using the formula,

So, substituting the values in the above mentioned formula

Thus,

Therefore, the number of terms present in the given A.P is

#### Question 6:

The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.

In the given problem, we are given 6th and 17th term of an A.P.

We need to find the 40th term

Here,

Now, we will find and using the formula

So,

Also,

So, to solve for a and d

On subtracting (1) from (2), we get

Substituting (3) in (1), we get

Thus,

Substituting the above values in the formula

Therefore,

#### Question 7:

The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term.

In the given problem, we are given 10th and 18th term of an A.P.

We need to find the 26th term

Here,

Now, we will find and using the formula

So,

Also,

So, to solve for a and d

On subtracting (1) from (2), we get

Substituting d=4 in (1), we get

Thus,

Substituting the above values in the formula,

Therefore,

#### Question 8:

If the nth term of the A.P. 9, 7, 5, ... is same as the nth term of the A.P. 15, 12, 9, ... find n.

Here, we are given two A.P. sequences whose nth terms are equal. We need to find n.

So let us first find the nth term for both the A.P.

First A.P. is 9, 7, 5 …

Here,

First term (a) = 9

Common difference of the A.P. (d)

Now, as we know,

So, for nth term,

Second A.P. is 15, 12, 9 …

Here,

First term (a) = 15

Common difference of the A.P. (d)

Now, as we know,

So, for nth term,

Now, we are given that the nth terms for both the A.P. sequences are equal, we equate (1) and (2),

Therefore,

#### Question 9:

Find the 12th term from the end of the following arithmetic progressions:

(i) 3, 5, 7, 9, ... 201

(ii) 3, 8, 13, ..., 253

(iii) 1, 4, 7, 10, ..., 88

In the given problem, we need to find the 12th term from the end for the given A.P.

(i) 3, 5, 7, 9 …201

Here, to find the 12th term from the end let us first find the total number of terms. Let us take the total number of terms as n.

So,

First term (a) = 3

Last term (an) = 201

Common difference (d) =

Now, as we know,

So, for the last term,

Further simplifying,

So, the 12th term from the end means the 89th term from the beginning.

So, for the 89th term (n = 89)

Therefore, the 12th term from the end of the given A.P. is.

(ii) 3, 8, 13 …253

Here, to find the 12th term from the end let us first find the total number of terms. Let us take the total number of terms as n.

So,

First term (a) = 3

Last term (an) = 253

Common difference, d =

Now, as we know,

So, for the last term,

Further simplifying,

So, the 12th term from the end means the 40th term from the beginning.

So, for the 40th term (n = 40)

Therefore, the 12th term from the end of the given A.P. is.

(iii) 1, 4, 7, 10 …88

Here, to find the 12th term from the end let us first find the total number of terms. Let us take the total number of terms as n.

So,

First term (a) = 1

Last term (an) = 88

Common difference, d= =3

Now, as we know,

So, for the last term,

Further simplifying,

So, the 12th term from the end means the 19th term from the beginning.

So, for the 19th term (n = 19)

Therefore, the 12th term from the end of the given A.P. is.

#### Question 10:

If 9th term of an A.P is zero, prove that its 29th term is double the 19th term.

In the given problem, the 9th term of an A.P. is zero.

Here, let us take the first term of the A.P as a and the common difference as d

So, as we know,

We get,

Now, we need to prove that 29th term is double of 19th term. So, let us first find the two terms.

For 19th term (n = 19),

(Using 1)

For 29th term (n = 29),

${a}_{29}=a+\left(29-1\right)d\phantom{\rule{0ex}{0ex}}=-8d+28d\phantom{\rule{0ex}{0ex}}=20d\phantom{\rule{0ex}{0ex}}=2×10d\phantom{\rule{0ex}{0ex}}=2×{a}_{19}$  (Using 1)

Therefore, for the given A.P. the 29th term is double of the 19th term.

Hence proved.

#### Question 11:

If 10 times the 10th term of an A.P. is equal to 15 times the 15th term, show that 25th term of the A.P. is zero.

Here, let us take the first term of the A.P. as a and the common difference as d

We are given that 10 times the 10th term is equal to 15 times the 15th term. We need to show that 25th term is zero.

So, let us first find the two terms.

So, as we know,

For 10th term (n = 10),

For 15th term (n = 15),

Now, we are given,

Solving this, we get,

Next, we need to prove that the 25th term of the A.P. is zero. For that, let us find the 25th term using n = 25,

Thus, the 25th term of the given A.P. is zero.

Hence proved

#### Question 12:

In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.

Here, we are given that 24th term is twice the 10th term, for a certain A.P. Here, let us take the first term of the A.P. as a and the common difference as d

We have to prove that

So, let us first find the two terms.

As we know,

For 10th term (n = 10),

For 24th term (n = 24),

Now, we are given that

So, we get,

Further, we need to prove that the 72nd term is twice of 34th term. So let now find these two terms,

For 34th term (n = 34),

(Using 1)

For 72nd term (n = 72),

(Using 1)

Therefore,

#### Question 13:

The 26th , 11th  and last term of an A.P. are 0, 3 and $-\frac{1}{5}$, respectively. Find the common difference and the  number of terms .

It is given that, .

#### Question 14:

The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1. Find the first term and the common difference.

In the given problem, let us take the first term as a and the common difference as d

Here, we are given that,

We need to find a and d

So, as we know,

For the 4th term (n = 4),

Similarly, for the 3rd term (n = 3),

Also, for the 7th term (n = 7),

Now, using the value of a3 in equation (2), we get,

Equating (3) and (4), we get,

On further simplification, we get,

Now, to find a,

Therefore, for the given A.P

#### Question 15:

Find the second term and nth term of an A.P. whose 6th term is 12 and 8th term is 22.

In the given problem, we are given 6th and 8th term of an A.P.

We need to find the 2nd and nth term

Here, let us take the first term as a and the common difference as d

We are given,

Now, we will find and using the formula

So,

Also,

So, to solve for a and d

On subtracting (1) from (2), we get

Substituting (3) in (1), we get

Thus,

So, for the 2nd term (n = 2),

For the nth term,

Therefore,

#### Question 16:

How many numbers of two digit are divisible by 3?

In this problem, we need to find out how many numbers of two digits are divisible by 3.

So, we know that the first two digit number that is divisible by 3 is 12 and the last two digit number divisible by 3 is 99. Also, all the terms which are divisible by 3 will form an A.P. with the common difference of 3.

So here,

First term (a) = 12

Last term (an) = 99

Common difference (d) = 3

So, let us take the number of terms as n

Now, as we know,

So, for the last term,

Further simplifying,

Therefore, the number of two digit terms divisible by 3 is.

#### Question 17:

An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find the 32nd term.

In the given problem, we need to find the 32nd term of an A.P. which contains a total of 60 terms.

Here we are given the following,

First term (a) = 7

Last term (an) = 125

Number of terms (n) = 60

So, let us take the common difference as d

Now, as we know,

So, for the last term,

Further simplifying,

So, for the 32nd term (n = 32)

Therefore, the 32nd term of the given A.P. is.

#### Question 18:

The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34. Find the first term and the common difference of the A.P.

In the given problem, the sum of 4th and 8th term is 24 and the sum of 6th and 10th term is 34.

We can write this as,

We need to find a and d

For the given A.P., let us take the first term as a and the common difference as d

As we know,

For 4th term (n = 4),

For 8th term (n = 8),

So, on substituting the above values in (1), we get,

Also, for 6th term (n = 6),

For 10th term (n = 10),

So, on substituting the above values in (2), we get,

Next we simplify (3) and (4). On subtracting (3) from (4), we get,

Further, using the value of d in equation (3), we get,

On further simplifying, we get,

Therefore, for the given A.P

#### Question 19:

The first term of an A.P. is 5 and its 100th term is −292. Find the 50th term of this A.P.

In the given problem, we are given 1st and 100th term of an A.P.

We need to find the 50th term

Here,

Now, we will find d using the formula

So,

Also,

So, to solve for d

Substituting a = 5, we get

Thus,

Substituting the above values in the formula,

Therefore,

#### Question 20:

Find a30 − a20 for the A.P.

(i) −9, −14, −19, −24, ...

(ii) a,a + d, a + 2d, a + 3d, ...

In this problem, we are given different A.P. and we need to find.

(i) A.P.

Here,

First term (a) = -9

Common difference of the A.P. (d)

Now, as we know,

Here, we find and

So, for 30th term,

Also, for 20th term,

So,

Therefore, for the given A.P

(ii) A.P.

Here,

First term (a) = a

Common difference of the A.P. (d) =d

Now, as we know,

Here, we find a30 and a20.

So, for 30th term,

Also, for 20th term,

So,

Therefore, for the given A.P

#### Question 21:

Write the expression an- ak for the A.P. a, a + d, a + 2d, ...
Hence, find the common difference of the A.P. for which

(i) 11th term is 5 and 13th term is 79.

(ii) a10 −a5 = 200

(iii) 20th term is 10 more than the 18th term.

A.P: a, a+d, a+2d …

Here, we first need to write the expression for

Now, as we know,

So, for nth term,

Similarly, for kth term

So,

So,

(i) In the given problem, we are given 11th and 13th term of an A.P.

We need to find the common difference. Let us take the common difference as d and the first term as a.

Here,

Now, we will find and using the formula

So,

Also,

Solving for a and d

On subtracting (1) from (2), we get

Therefore, the common difference for the A.P. is.

(ii) We are given,

Here,

Let us take the first term as a and the common difference as d

Now, as we know,

Here, we find a30 and a20.

So, for 10th term,

Also, for 5th term,

So,

Therefore, the common difference for the A.P. is.

(iii) In the given problem, the 20th term is 10 more than the 18th term. So, let us first find the 20th term and 18th term of the A.P.

Here

Let us take the first term as a and the common difference as d

Now, as we know,

So, for 20th term (n = 20),

Also, for 18th term (n = 18),

Now, we are given,

On substituting the values, we get,

Therefore, the common difference for the A.P. is

#### Question 22:

Find n if the given value of x is the nth term of the given A.P.

(i) 25, 50, 75, 100, ...; x = 1000

(ii) −1, −3, −5, −7, ...; x = −151

(iii)

(iv)

In the given problem, we need to find the number of terms in an A.P

(i) 25, 50, 75, 100 …

We are given,

Let us take the total number of terms as n.

So,

First term (a) = 25

Last term (an) = 1000

Common difference (d) =

Now, as we know,

So, for the last term,

Therefore, the total number of terms of the given A.P. is.

(ii) -1, -3, -5, -7 …

We are given,

Let us take the total number of terms as n.

So,

First term (a) = −1

Last term (an) = −151

Common difference (d) =

Now, as we know,

So, for the last term,

On further simplifying, we get,

Therefore, the total number of terms of the given A.P. is.

(iii)

We are given,

Let us take the total number of terms as n.

So,

First term (a) =

Last term (an) = 550

Common difference (d) =

Now, as we know,

So, for the last term,

On further simplifying, we get,

Therefore, the total number of terms of the given A.P. is

(iv)

We are given,

Let us take the total number of terms as n.

So,

First term (a) = 1

Last term (an) =

Common difference (d) =

Now, as we know,

So, for the last term,

On further simplifying, we get,

Therefore, the total number of terms of the given A.P. is.

#### Question 23:

THe eighth term of an A.P. is half of its second term and the eleventh term exceeds one third of its fourth term by 1 . Find the 15th term.

#### Question 24:

Find the arithmetic progression whose third term is 16 and seventh term exceeds its fifth term by 12.

Here, let us take the first term of the A.P as a and the common difference of the A.P as d

Now, as we know,

So, for 3rd term (n = 3),

Also, for 5th term (n = 5),

For 7th term (n = 7),

Now, we are given,

Substituting the value of d in (1), we get,

So, the first term is 4 and the common difference is 6.

Therefore, the A.P. is

#### Question 25:

The 7th term of an A.P. is 32 and its 13th term is 62.  Find the A.P.

Here, let us take the first term of the A.P. as a and the common difference of the A.P as d

Now, as we know,

So, for 7th term (n = 7),

Also, for 13th term (n = 13),

Now, on subtracting (2) from (1), we get,

Substituting the value of d in (1), we get,

So, the first term is 2 and the common difference is 5.

Therefore, the A.P. is

#### Question 26:

Which term of the A.P. 3, 10, 17, ... will be 84 more than its 13th term?

In the given problem, let us first find the 13th term of the given A.P.

A.P. is 3, 10, 17 …

Here,

First term (a) = 3

Common difference of the A.P. (d) =7

Now, as we know,

So, for 13th term (n = 13),

Let us take the term which is 84 more than the 13th term as an. So,

Also,

Further simplifying, we get,

Therefore, theof the given A.P. is 84 more than the 13th term.

#### Question 27:

Two arithmetic progression have the same common difference. The difference between their 100th terms is 100, What is the difference between their 1000th terms?

Here, we are given two A.P sequences which have the same common difference. Let us take the first term of one A.P. as a and of other A.P. as a’

Also, it is given that the difference between their 100th terms is 100.

We need to find the difference between their 100th terms

So, let us first find the 100th terms for both of them.

Now, as we know,

So, for 100th term of first A.P. (n = 100),

Now, for 100th term of second A.P. (n = 100),

Now, we are given,

On substituting the values, we get,

Now, we need the difference between the 1000th terms of both the A.P.s

So, for 1000th term of first A.P. (n = 1000),

Now, for 1000th term of second A.P. (n = 1000),

So,

Therefore, the difference between the 1000th terms of both the arithmetic progressions will be.

#### Question 28:

For what value of n, the nth terms of the arithmetic progressions 63, 65, 67, ... and 3, 10, 17, ... are equal?

Here, we are given two A.P. sequences. We need to find the value of n for which the nth terms of both the sequences are equal. We need to find n

So let us first find the nth term for both the A.P.

First A.P. is 63, 65, 67 …

Here,

First term (a) = 63

Common difference of the A.P. (d) =2

Now, as we know,

So, for nth term,

Second A.P. is 3, 10, 17 …

Here,

First term (a) = 3

Common difference of the A.P. (d) =7

Now, as we know,

So, for nth term,

Now, we are given that the nth terms for both the A.P. sequences are equal, we equate (1) and (2),

Therefore,

#### Question 29:

How many multiples of 4 lie between 10 and 250?

In this problem, we need to find out how many multiples of 4 lie between 10 and 250.

So, we know that the first multiple of 4 after 10 is 12 and the last multiple of 4 before 250 is 248. Also, all the terms which are divisible by 4 will form an A.P. with the common difference of 4.

So here,

First term (a) = 12

Last term (an) = 248

Common difference (d) = 4

So, let us take the number of terms as n

Now, as we know,

So, for the last term,

Further simplifying,

Therefore, the number of multiples of 4 that lie between 10 and 250 is.

#### Question 30:

How many three digit numbers are divisible by 7?

In this problem, we need to find out how many numbers of three digits are divisible by 7.

So, we know that the first three digit number that is divisible by 7 is 105 and the last three digit number divisible by 7 is 994. Also, all the terms which are divisible by 7 will form an A.P. with the common difference of 7.

So here,

First term (a) = 105

Last term (an) = 994

Common difference (d) = 7

So, let us take the number of terms as n

Now, as we know,

So, for the last term,

Further simplifying,

Therefore, the number of three digit terms divisible by 7 is.

#### Question 31:

Which term of the arithmetic progression 8, 14, 20, 26, ... will be 72 more than its 41st term.

In the given problem, let us first find the 41st term of the given A.P.

A.P. is 8, 14, 20, 26 …

Here,

First term (a) = 8

Common difference of the A.P. (d) =6

Now, as we know,

So, for 41st term (n = 41),

Let us take the term which is 72 more than the 41st term as an. So,

Also,

Further simplifying, we get,

Therefore, the of the given A.P. is 72 more than the 41st term.

#### Question 32:

Find the term of the arithmetic progression 9, 12, 15, 18, ... which is 39 more than its 36th term.

In the given problem, let us first find the 36st term of the given A.P.

A.P. is 9, 12, 15, 18 …

Here,

First term (a) = 9

Common difference of the A.P. (d) =3

Now, as we know,

So, for 36th term (n = 36),

Let us take the term which is 39 more than the 36th term as an. So,

Also,

Further simplifying, we get,

Therefore, theof the given A.P. is 39 more than the 36th term

#### Question 33:

Find the 8th term from the end of the A.P. 7, 10, 13, ..., 184.

In the given problem, we need to find the 8th term from the end for the given A.P.

We have the A.P as 7, 10, 13 …184

Here, to find the 8th term from the end let us first find the total number of terms. Let us take the total number of terms as n.

So,

First term (a) = 7

Last term (an) = 184

Common difference (d) = =3

Now, as we know,

So, for the last term,

Further simplifying,

So, the 8th term from the end means the 53rd term from the beginning.

So, for the 53rd term (n = 53)

Therefore, the 8th term from the end of the given A.P. is.

#### Question 34:

Find the 10th term from the end of the A.P. 8, 10, 12, ..., 126.

In the given problem, we need to find the 10th term from the end for the given A.P.

We have the A.P as 8, 10, 12 …126

Here, to find the 10th term from the end let us first find the total number of terms. Let us take the total number of terms as n.

So,

First term (a) = 8

Last term (an) = 126

Common difference (d) = =2

Now, as we know,

So, for the last term,

Further simplifying,

So, the 10th term from the end means the 51st term from the beginning.

So, for the 51st term (n = 51)

Therefore, the 10th term from the end of the given A.P. is.

#### Question 35:

The sum 4th and 8th terms of an A.P. is 24 and the sum of 6th and 10th terms is 44. Find the A.P.

In the given problem, the sum of 4th and 8th term is 24 and the sum of 6th and 10th term is 44. We have to find the A.P

We can write this as,

We need to find the A.P

For the given A.P., let us take the first term as a and the common difference as d

As we know,

For 4th term (n = 4),

For 8th term (n = 8),

So, on substituting the above values in (1), we get,

Also, for 6th term (n = 6),

For 10th term (n = 10),

So, on substituting the above values in (2), we get,

Next we simplify (3) and (4). On subtracting (3) from (4), we get,

Further, using the value of d in equation (3), we get,

So here,

Therefore, the A.P. is.

#### Question 36:

Which term of the A.P. 3, 15, 27, 39, ... will be 120 more than its 21st term?

In the given problem, let us first find the 21st term of the given A.P.

A.P. is 3, 15, 27, 39 …

Here,

First term (a) = 3

Common difference of the A.P. (d) =12

Now, as we know,

So, for 21st term (n = 21),

Let us take the term which is 120 more than the 21st term as an. So,

Also,

Further simplifying, we get,

Therefore, theof the given A.P. is 120 more than the 21st term.

#### Question 37:

The 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43. find the nth term.

Let a be the first term and d be the common difference.

We know that, nth term = an = a + (n − 1)d

According to the question,

a17 = 5 + 2a8
a + (17 − 1)d =  5 + 2(a + (8 − 1)d)
⇒ a + 16d =  5 + 2a + 14d
⇒ 16d − 14=  5 + 2a − a
⇒ 2=  5 + a
a = 2− 5    .... (1)

Also, a11 = 43
⇒ a + (11 − 1)d = 43
⇒ a + 10d = 43   ....(2)

On substituting the values of (1) in (2), we get
2− 5 + 10d = 43
⇒ 12= 5 + 43
⇒ 12= 48
⇒ = 4
⇒ a = 2 × 4 − 5     [From (1)]
⇒ a = 3

∴ an a + (n − 1)d
=
3 + (n − 1)4
= 3 + 4n − 4
= 4n − 1

Thus, the nth term of the given A.P. is  4n − 1.

#### Question 38:

Find the number of all three digit natural numbers which are divisible by 9.

First three-digit number that is divisible by 9 is 108.

Next number is 108 + 9 = 117.

And the last three-digit number that is divisible by 9 is 999.

Thus, the progression will be 108, 117, .... , 999.

All are three digit numbers which are divisible by 9, and thus forms an A.P. having first term a 108 and the common difference as 9.

We know that, nth term = an a + (n − 1)d

According to the question,

999 = 108 + (n − 1)9
⇒ 108 + 9n − 9 = 999
⇒ 99 + 9n = 999
⇒ 9n = 999 − 99
⇒ 9n = 900
⇒ n = 100

Thus, the number of all three digit natural numbers which are divisible by 9 is 100.

#### Question 39:

The 19th term of an A.P. is equal to three times its sixth term. If its 9th term is 19, find the A.P.

Let a be the first term and d be the common difference.

We know that, nth term = an a + (n − 1)d

According to the question,

a19 = 3a6
⇒ a + (19 − 1)d =  3(a + (6 − 1)d)
⇒ a + 18d =  3a + 15d
⇒ 18d − 15=  3a − a
⇒ 3= 2a
⇒ a = $\frac{3}{2}$d   .... (1)

Also, a9 = 19
⇒ a + (9 − 1)d = 19
⇒ a + 8d = 19   ....(2)

On substituting the values of (1) in (2), we get
$\frac{3}{2}$+ 8d = 19
⇒ 3+ 16= 19 × 2
⇒ 19= 38
⇒ = 2
⇒ a = $\frac{3}{2}×2$     [From (1)]
⇒ a = 3

Thus, the A.P. is 3, 5, 7, 9, .... .

#### Question 40:

The 9th term of an A.P. is equal to 6 times its second term. If its 5th term is 22, find the A.P.

Let a be the first term and d be the common difference.

We know that, nth term = an a + (n − 1)d

According to the question,

a9 = 6a2
⇒ a + (9 − 1)d =  6(a + (2 − 1)d)
⇒ a + 8d =  6a + 6d
⇒ 8d − 6=  6a − a
⇒ 2= 5a
⇒ a = $\frac{2}{5}$d   .... (1)

Also, a5 = 22
⇒ a + (5 − 1)d = 22
⇒ a + 4d = 22   ....(2)

On substituting the values of (1) in (2), we get
$\frac{2}{5}$+ 4d = 22
⇒ 2+ 20= 22 × 5
⇒ 22= 110
⇒ = 5
⇒ a = $\frac{2}{5}×5$     [From (1)]
⇒ a = 2

Thus, the A.P. is 2, 7, 12, 17, .... .

#### Question 41:

The 24th term of an A.P. is twice its 10th term. Show that its 72nd term is 4 times its 15th term.

Let a be the first term and d be the common difference.

We know that, nth term = an a + (n − 1)d

According to the question,

a24 = 2a10
⇒ a + (24 − 1)d = 2(a + (10 − 1)d)
⇒ a + 23d = 2a + 18d
⇒ 23d − 18= 2a − a
⇒ 5= a
⇒ a = 5d   .... (1)

Also,
a72 = a + (72 − 1)d
= 5d + 71d              [From (1)]
= 76d                      ..... (2)

and
a15 a + (15 − 1)d
= 5d + 14d              [From (1)]
= 19d                      ..... (3)

On comparing (2) and (3), we get

76= 4 × 19d
⇒ a72 = 4 × a15

Thus, 72nd term of the given A.P. is 4 times its 15th term.

#### Question 42:

Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.

Since, the number is divisible by both 2 and 5, means it must be divisible by 10.

In the given numbers, first number that is divisible by 10 is 110.

Next number is 110 + 10 = 120.

The last number that is divisible by 10 is 990.

Thus, the progression will be 110, 120, ..., 990.

All the terms are divisible by 10, and thus forms an A.P. having first term as 110 and the common difference as 10.

We know that, nth term = an a + (n − 1)d

According to the question,

990 = 110 + (n − 1)10
⇒ 990 = 110 + 10n − 10
⇒ 10n = 990 − 100
⇒ 10= 890
⇒ n = 89

Thus, ​the number of natural numbers between 101 and 999 which are divisible by both 2 and 5 is 89.

#### Question 43:

If the seventh term of an A.P. is $\frac{1}{9}$ and its ninth term is $\frac{1}{7}$, find its (63)rd term.

Let a be the first term and d be the common difference.

We know that, nth term = an a + (n − 1)d

According to the question,

a7 = $\frac{1}{9}$
⇒ a + (7 − 1)d = $\frac{1}{9}$
⇒ a + 6d = $\frac{1}{9}$       .... (1)

Also, a9 = $\frac{1}{7}$
⇒ a + (9 − 1)d = $\frac{1}{7}$
⇒ a + 8d = $\frac{1}{7}$   ....(2)

On Subtracting (1) from (2), we get
8− 6d = $\frac{1}{7}-\frac{1}{9}$
⇒ 2$\frac{9-7}{63}$
⇒ 2= $\frac{2}{63}$
⇒ $\frac{1}{63}$
⇒ a = $\frac{1}{9}-\frac{6}{63}$     [From (1)]
⇒ a = $\frac{7-6}{63}$
⇒ a = $\frac{1}{63}$

∴ a63 a + (63 − 1)d
=
$\frac{1}{63}+\frac{62}{63}$
= $\frac{63}{63}$
= 1

Thus, (63)rd term of the given A.P. is 1.

#### Question 44:

The sum of 5th and 9th terms of an A.P. is 30. If its 25th term is three times its 8th term, find the A.P.

Let a be the first term and d be the common difference.

We know that, nth term = an a + (n − 1)d

According to the question,

a5 + a9 = 30
⇒ a + (5 − 1)d + a + (9 − 1)= 30
⇒ a + 4d + a + 8d = 30
⇒ 2a + 12d = 30
a + 6d = 15       .... (1)

Also, a25 = 3(a8)
⇒ a + (25 − 1)d = 3[a + (8 − 1)d]
⇒ a + 24d = 3a + 21d
⇒ 3a − a = 24d − 21d
⇒ 2a = 3d
a = $\frac{3}{2}$d   ....(2)

Substituting the value of (2) in (1), we get
$\frac{3}{2}$+ 6d = 15
⇒ 3+ 12= 15 × 2
⇒ 15= 30
⇒ = 2
⇒ a = $\frac{3}{2}×2$         [From (1)]
⇒ a = 3

Thus, the A.P. is 3, 5, 7, 9, .... .

#### Question 45:

Find where 0 (zero) is a term of the A.P. 40, 37, 34, 31, ..... .

Let a be the first term and d be the common difference.

We know that, nth term = an a + (n − 1)d

It is given that a = 40, d = −3 and an = 0

According to the question,

⇒ 0 = 40 + (n − 1)(−3)
⇒ 0 = 40 − 3n + 3
⇒ 3n = 43
⇒ n = $\frac{43}{3}$          .... (1)

Here, n is the number of terms, so must be an integer.

Thus, there is no term where 0 (zero) is a term of the A.P. 40, 37, 34, 31, ..... .

#### Question 46:

Find the middle term of the A.P. 213, 205, 197, ...., 37.

Let a be the first term and d be the common difference.

We know that, nth term = an a + (n − 1)d

It is given that a = 213, d = −8 and an = 37

According to the question,

⇒ 37 = 213 + (n − 1)(−8)
⇒ 37 = 213 − 8n + 8
⇒ 8n = 221 − 37
⇒ 8n = 184
⇒ n = 23          .... (1)

Therefore, total number of terms is 23.

Since, there are odd number of terms.
So, Middle term will be ${\left(\frac{23+1}{2}\right)}^{\mathrm{th}}$ term, i.e., the 12th term.

∴ a12 = 213 + (12 − 1)(−8)
= 213 − 88
= 125

Thus, the middle term of the A.P. 213, 205, 197, ...., 37 is 125.

#### Question 47:

If the 5th term of an A.P. is 31 and 25th term is 140 more than the 5th term, find the A.P.

Let a be the first term and d be the common difference.

We know that, nth term = an a + (n − 1)d

According to the question,

a5 = 31
⇒ a + (5 − 1)d =  31
⇒ a + 4d =  31
⇒ a = 31 − 4d        .... (1)

Also, a25 = 140 + a5
⇒ a + (25 − 1)d = 140 + 31
⇒ a + 24d = 171    .... (2)

On substituting the values of (1) in (2), we get
31 − 4d + 24d = 171
⇒ 20= 171 − 31
⇒ 20= 140
⇒ = 7
⇒ a = 31 − 4 × 7     [From (1)]
⇒ a = 3

Thus, the A.P. is 3, 10, 17, 24, .... .

#### Question 48:

Find the sum of two middle terms of the A.P. : .

#### Question 49:

If (m + 1)th term of an A.P is twice the (n + 1)th term, prove that (3m + 1)th term is twice the (m + n + 1)th term.

Here, we are given that (m+1)th term is twice the (n+1)th term, for a certain A.P. Here, let us take the first term of the A.P. as a and the common difference as d

We need to prove that

So, let us first find the two terms.

As we know,

For (m+1)th term (n’ = m+1)

For (n+1)th term (n’ = n+1),

Now, we are given that

So, we get,

Further, we need to prove that the (3m+1)th term is twice of (m+n+1)th term. So let us now find these two terms,

For (m+n+1)th term (n’ = m+n+1 ),

(Using 1)

For (3m+1)th term (n’ = 3m+1),

Therefore,

Hence proved

#### Question 50:

If an A.P. consists of n terms with first term a and nth term l show that the sum of the mth term from the beginning and the mth term from the end is (a + l).

In the given problem, we have an A.P. which consists of n terms.

Here,

The first term (a) = a

The last term (an) = l

Now, as we know,

So, for the mth term from the beginning, we take (n = m),

Similarly, for the mth term from the end, we can take l as the first term.

So, we get,

Now, we need to prove

So, adding (1) and (2), we get,

Therefore,

Hence proved

#### Question 51:

How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3 ?

The given A.P is 11, 15, 19, .....,299.
$a=11\phantom{\rule{0ex}{0ex}}d=4\phantom{\rule{0ex}{0ex}}{a}_{n}=299\phantom{\rule{0ex}{0ex}}⇒a+\left(n-1\right)d=299\phantom{\rule{0ex}{0ex}}⇒11+\left(n-1\right)4=299\phantom{\rule{0ex}{0ex}}⇒\left(n-1\right)4=288\phantom{\rule{0ex}{0ex}}⇒n=73$

#### Question 52:

Find the 12th term from the end of the A.P.  .

Consider the A.P −2, −4, −6, ...., −100.

#### Question 53:

For the A.P. : can we find  withoput actually finding   ? Give reasons for your answer.

Consider the A.P −3, −7, −11, ....

#### Question 54:

Two A.P.s have the same common difference . The first term of one A.P. is 2 and that of the other is 7 . The difference between their 10th terms is the same as the difference between their 21st terms , which is the same as the difference between any two corresponding terms . Why ?

First term of 1st A.P is 2.
First term of 2nd A.P is 7.
Consider the difference of their 10th terms,

The difference between any two corresponding terms of A.P's is same as the difference between their terms.

#### Question 1:

Find the value of x for which (8x + 4), (6x − 2) and (2x + 7) are in A.P.

Here, we are given three terms,

First term (a1) =

Second term (a2) =

Third term (a3) =

We need to find the value of x for which these terms are in A.P. So, in an A.P. the difference of two adjacent terms is always constant. So, we get,

…… (1)

Also,

…… (2)

Now, on equating (1) and (2), we get,

Therefore, for, these three terms will form an A.P.

#### Question 2:

If x + 1, 3x and 4x + 2 are in A.P., find the value of x.

Here, we are given three terms which are in A.P.,

First term (a1) =

Second term (a2) =

Third term (a3) =

We need to find the value of x. So, in an A.P. the difference of two adjacent terms is always constant. So, we get,

…… (1)

Also,

…… (2)

Now, on equating (1) and (2), we get,

Therefore, for, these three terms will form an A.P.

#### Question 3:

Show that (a − b)2, (a2 + b2) and (a + b)2 are in A.P.

Here, we are given three terms and we need to show that they are in A.P.,

First term (a1) =

Second term (a2) =

Third term (a3) =

So, in an A.P. the difference of two adjacent terms is always constant. So, to prove that these terms are in A.P. we find the common difference, we get,

…… (1)

Also,

…… (2)

Now, since in equations (1) and (2) the values of d are equal, we can say that these terms are in A.P. with 2ab as the common difference.

Hence proved

#### Question 4:

Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers.

In the given problem, the sum of three terms of an A.P is 27 and the product of the three terms is 648. We need to find the three terms.

Here,

Let the three terms be where a is the first term and d is the common difference of the A.P

So,

…… (1)

Also,

Further solving for d,

…… (2)

Now, substituting (1) and (2) in three terms

First term =

So,

Also,

Second term = a

So,

Also,

Third term =

So,

Therefore, the three terms are .

#### Question 5:

The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.

In the given problem, the sum of three terms of an A.P is 21 and the product of the first and the third term exceeds the second term by 6.

We need to find the three terms.

Here,

Let the three terms be where, a is the first term and d is the common difference of the A.P

So,

…… (1)

Also,

(Using)

(Using 1)

Further solving for d,

…… (2)

Now, using the values of a and d in the expressions of the three terms, we get,

First term =

So,

Second term = a

So,

Also,

Third term =

So,

Therefore, the three terms are .

#### Question 6:

Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.

Here, we are given that four number are in A.P., such that there sum is 50 and the greatest number is 4 times the smallest.

So, let us take the four terms as.

Now, we are given that sum of these numbers is 50, so we get,

…… (1)

Also, the greatest number is 4 times the smallest, so we get,

…… (2)

Now, using (2) in (1), we get,

Now, using the value of a in (2), we get

So, first term is given by,

Second term is given by,

Third term is given by,

Fourth term is given by,

Therefore, the four terms are .

#### Question 7:

The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. Find the numbers.

In the given problem, the sum of three terms of an A.P is 12 and the sum of their cubes is 288.

We need to find the three terms.

Here,

Let the three terms be where, a is the first term and d is the common difference of the A.P

So,

Also, it is given that

So, using the properties:

We get,

Further solving for d by substituting the value of a, we get,

On further simplification, we get,

Now, here d can have two values +2 and -2.

So, on substituting the values of a = 4 and d = 2 in three terms, we get,

First term =

So,

Second term = a

So,

Third term =

So,

Also, on substituting the values of a = 4 and in three terms, we get,

First term =

So,

Second term = a

So,

Third term =

So,

Therefore, the three terms are .

#### Question 8:

Divide 56 in four parts in A.P. such that the ratio of the product of their extremes to the product of their means is 5 : 6.

Let the four terms of the AP be a − 3d, a − d, a + d and a + 3d.
Given:
(− 3d) + (a − d) + (a + d) + (a + 3d) = 56
$⇒$ 4a = 56
$⇒$ a = 14

$⇒1176-54{d}^{2}=980-5{d}^{2}\phantom{\rule{0ex}{0ex}}⇒196=49{d}^{2}\phantom{\rule{0ex}{0ex}}⇒{d}^{2}=4\phantom{\rule{0ex}{0ex}}⇒d=±2$

When d = 2, the terms of the AP are 8, 12, 16, 20. When d−2, the terms of the AP are 20, 18, 12, 8.

#### Question 9:

The sum of the first three numbers in an arithmetic progression is 18. If the product of the first and third term is 5 times the common difference, find the three numbers.

Let the first three numbers in an arithmetic progression be a − d, a, a + d.

The sum of the first three numbers in an arithmetic progression is 18.

a − d + a + a + d = 18
⇒ 3a = 18
a = 6

The product of the first and third term is 5 times the common difference.

Hence, the three numbers are 2, 6, 10 or 15, 6, −3.

#### Question 10:

The angles of a quadrilateral are in A.P. whose common difference is 10°. Find the numbers.

Here, we are given that the angles of a quadrilateral are in A.P, such that the common difference is 10°.

So, let us take the angles as

Now, we know that the sum of all angles of a quadrilateral is 360°. So, we get,

On further simplifying for a, we get,

So, the first angle is given by,

Second angle is given by,

Third angle is given by,

Fourth angle is given by,

Therefore, the four angles of the quadrilateral are .

#### Question 11:

Split  207 into three parts such that these are in A.P. and the product of the two smaller parts is 4623.

Suppose three parts of 207 are (a d), a , (a + d) such that , (a + d)  >a >  (a − d).

#### Question 12:

The angles of a triangle are in A.P. The greatest angle is twice the least . Find all the angles.

Suppose the angles of a triangle are (a d), a , (a + d) such that , (a + d)  >a >  (a − d).

#### Question 13:

The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7 : 15 . Find the number .

Let the four terms of the AP be a − 3d, a − d, a + d and a + 3d.
Given:
(− 3d) + (a − d) + (a + d) + (a + 3d) = 32
$⇒$ 4a = 32
$⇒$ a = 8

$⇒960-135{d}^{2}=448-7{d}^{2}\phantom{\rule{0ex}{0ex}}⇒512=128{d}^{2}\phantom{\rule{0ex}{0ex}}⇒{d}^{2}=4\phantom{\rule{0ex}{0ex}}⇒d=±2$
When a = 8 and d= 2, then the terms are 2, 6, 10, 14.
When a = 8 and d= −2, then the terms are 14, 10, 6, 2.

#### Question 1:

Find the sum of the following arithmetic progressions:

(i) 50, 46, 42, ... to 10 terms

(ii) 1, 3, 5, 7, ... to 12 terms

(iii) 3, 9/2, 6, 15/2, ... to 25 terms

(iv) 41, 36, 31, ... to 12 terms

(v) a + b, a − b, a − 3b, ... to 22 terms

(vi) (x − y)2, (x2 + y2), (x + y)2, ..., to n terms

(vii) to n terms

(viii) −26, −24, −22, ... to 36 terms.

In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of n terms of an A.P.,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

(i) To 10 terms

Common difference of the A.P. (d)

=

Number of terms (n) = 10

First term for the given A.P. (a) = 50

So, using the formula we get,

Therefore, the sum of first 10 terms for the given A.P. is.

(ii) To 12 terms.

Common difference of the A.P. (d)

=

Number of terms (n) = 12

First term for the given A.P. (a) = 1

So, using the formula we get,

Therefore, the sum of first 12 terms for the given A.P. is.

(iii) To 25 terms.

Common difference of the A.P. (d) =

Number of terms (n) = 25

First term for the given A.P. (a) = 3

So, using the formula we get,

On further simplifying, we get,

Therefore, the sum of first 25 terms for the given A.P. is.

(iv) To 12 terms.

Common difference of the A.P. (d) =

Number of terms (n) = 12

First term for the given A.P. (a) = 41

So, using the formula we get,

Therefore, the sum of first 12 terms for the given A.P. is.

(v) To 22 terms.

Common difference of the A.P. (d) =

Number of terms (n) = 22

First term for the given A.P. (a) =

So, using the formula we get,

Therefore, the sum of first 22 terms for the given A.P. is.

(vi) To n terms.

Common difference of the A.P. (d) =

Number of terms (n) = n

First term for the given A.P. (a) =

So, using the formula we get,

Now, taking 2 common from both the terms inside the bracket we get,

Therefore, the sum of first n terms for the given A.P. is

(vii) To n terms.

Number of terms (n) = n

First term for the given A.P. (a) =

Common difference of the A.P. (d) =

So, using the formula we get,

Now, on further solving the above equation we get,

Therefore, the sum of first n terms for the given A.P. is.

(viii) To 36 terms.

Common difference of the A.P. (d) =

Number of terms (n) = 36

First term for the given A.P. (a) = −26

So, using the formula we get,

Therefore, the sum of first 36 terms for the given A.P. is.

#### Question 1:

Write the first five terms of each of the following sequences whose nth terms are:

(a) an = 3n + 2

(b)

(c) an = 3n

(d)  ${a}_{n}=\frac{3n-2}{5}$

(e) an = (−1)n 2n

(f) ${a}_{n}=\frac{n\left(n-2\right)}{2}$

(g) an =n2 − n + 1

(h) an = 2n2 − 3n + 1

(i)

Here, we are given the nth term for various sequences. We need to find the first five terms of the sequence.

(i)

Here, the nth term is given by the above expression. So, to find the first term we use, we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms for the given sequence are.

(ii)

Here, the nth term is given by the above expression. So, to find the first term we use, , we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms for the given sequence are.

(iii)

Here, the nth term is given by the above expression. So, to find the first term we use, we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms for the given sequence are.

(iv)

Here, the nth term is given by the above expression. So, to find the first term we use, , we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms for the given sequence are.

(v)

Here, the nth term is given by the above expression. So, to find the first term we use, we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms of the given A.P are.

(vi)

Here, the nth term is given by the above expression. So, to find the first term we use , we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms for the given sequence are.

(vii)

Here, the nth term is given by the above expression. So, to find the first term we use, we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms for the given sequence are.

(viii)

Here, the nth term is given by the above expression. So, to find the first term we use, we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms for the given sequence are.

(ix)

Here, the nth term is given by the above expression. So, to find the first term we use, we get,

Similarly, we find the other four terms,

Second term (),

Third term (),

Fourth term (),

Fifth term (),

Therefore, the first five terms of the given A.P are

#### Question 2:

Find the indicated terms in each of the following sequences whose nth terms are:

(a) an = 5n − 4; a12 and a15

(b)

(c) an = n (n −1) (n − 2); a5 and a8

(d) an = (n − 1) (2 − n) (3 + n); a1, a2, a3

(e) an = (−1)n n ; a3, a5, a8

Here, we are given the nth term for various sequences. We need to find the indicated terms of the A.P.

(i)

We need to find and

Now, to find term we use, we get,

Also, to find term we use, we get,

${a}_{15}=5\left(15\right)-4\phantom{\rule{0ex}{0ex}}=75-4\phantom{\rule{0ex}{0ex}}=71$

Thus,

(ii)

We need to find and

Now, to find term we use, we get,

Also, to find term we use, we get,

Thus,

(iii)

We need to find and

Now, to find term we use, we get,

Also, to find term we use, we get,

Thus,

(iv)

We need to find, and

Now, to find term we use, we get,

Also, to find term we use, we get,

Similarly, to find term we use, we get,

Thus,

(v)

We need to find, and

Now, to find term we use, we get,

Also, to find term we use, we get,

Similarly, to find term we use, we get,

Thus,

#### Question 3:

Find the next five terms of each of the following sequences given by:

(i) a1 = 1, an = an−1+2, n ≥ 2

(ii) a1 = a2 = 2, an = an−1 − 3, n > 2

(iii)

(iv) a1 = 4, an = 4an−1 + 3, n > 1.

In the given problem, we are given the first, second term and the nth term of an A.P.

We need to find its next five terms

(i) , ,

Here, we are given that

So, the next five terms of this A.P would be,, , and

Now …… (1)

So, to find theterm we use, we get,

(Using 1)

…… (2)

For, using, we get,

(Using 2)

…… (3)

For, using, we get,

(Using 3)

…… (4)

For, using, we get,

(Using 4)

……. (5)

For, using, we get,

(Using 5)

Therefore, the next five terms, of the given A.P are

(ii) , ,

Here, we are given that

So, the next five terms of this A.P would be,,,and

Now …… (1)

So, to find theterm we use, we get,

(Using 1)

…… (2)

For, using, we get,

(Using 2)

…… (3)

For, using, we get,

(Using 3)

…… (4)

For, using, we get,

(Using 4)

…… (5)

For, using, we get,

(Using 5)

Therefore, the next five terms, of the given A.P are

(iii) , ,

Here, we are given that

So, the next five terms of this A.P would be, ,, and

Now …… (1)

So, to find theterm we use, we get,

(Using 1)

…… (2)

For, using, we get,

(Using 2)

…… (3)

For, using, we get,

(Using 3)

…… (4)

For, using, we get,

(Using 4)

…… (5)

For, using, we get,

(Using 5)

Therefore, the next five terms, of the given A.P are

(iv) , ,

Here, we are given that n > 1.

So, the next five terms of this A.P would be, , , and

Now …… (1)

So, to find theterm we use, we get,

(Using 1)

…… (2)

For, using, we get,

(Using 2)

…… (3)

For, using, we get,

(Using 3)

…… (4)

For, using, we get,

(Using 4)

…… (5)

For, using, we get,

(Using 5)

Therefore, the next five terms, of the given A.P are

#### Question 2:

Find the sum to n term of the A.P. 5, 2, −1, −4, −7, ...,

In the given problem, we need to find the sum of the n terms of the given A.P. “”.

So, here we use the following formula for the sum of n terms of an A.P.,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

For the given A.P. (),

Common difference of the A.P. (d) =

Number of terms (n) = n

First term for the given A.P. (a) = 5

So, using the formula we get,

Therefore, the sum of first n terms for the given A.P. is.

#### Question 3:

Find the sum of n terms of an A.P. whose nth terms is given by an = 5 − 6n.

Here, we are given an A.P., whose nth term is given by the following expression,

So, here we can find the sum of the n terms of the given A.P., using the formula,

Where, a = the first term

l = the last term

So, for the given A.P,

The first term (a) will be calculated using in the given equation for nth term of A.P.

Now, the last term (l) or the nth term is given

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

Therefore, the sum of the n terms of the given A.P. is.

#### Question 4:

Find the sum of last ten terms of the A.P. : 8, 10, 12, 14,...., 126.

The given A.P is 8, 10, 12, 14,...., 126.
a = 8 and d = 2.
When this A.P is reversed, we get the A.P.
126, 124, 122, 120,....
So, first term becomes 126 and common difference −2.
The sum of first 10 terms of this A.P is as follows:
${S}_{10}=\frac{10}{2}\left[2×126+9\left(-2\right)\right]\phantom{\rule{0ex}{0ex}}=5\left[234\right]\phantom{\rule{0ex}{0ex}}=1170$

#### Question 5:

Find the sum of the first 15 terms of each of the following sequences having nth term as

(i) an = 3 + 4n

(ii) bn = 5 + 2n

(iii) xn = 6 − n

(iv) yn = 9 − 5n

(i) Here, we are given an A.P. whose nth term is given by the following expression, . We need to find the sum of first 15 terms.

So, here we can find the sum of the n terms of the given A.P., using the formula,

Where, a = the first term

l = the last term

So, for the given A.P,

The first term (a) will be calculated using in the given equation for nth term of A.P.

Now, the last term (l) or the nth term is given

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

Therefore, the sum of the 15 terms of the given A.P. is.

(ii) Here, we are given an A.P. whose nth term is given by the following expression

We need to find the sum of first 15 terms.

So, here we can find the sum of the n terms of the given A.P., using the formula,

Where, a = the first term

l = the last term

So, for the given A.P,

The first term (a) will be calculated using in the given equation for nth term of A.P.

Now, the last term (l) or the nth term is given

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

Therefore, the sum of the 15 terms of the given A.P. is.

(iii) Here, we are given an A.P. whose nth term is given by the following expression, . We need to find the sum of first 15 terms.

So, here we can find the sum of the n terms of the given A.P., using the formula,

Where, a = the first term

l = the last term

So, for the given A.P,

The first term (a) will be calculated using in the given equation for nth term of A.P.

Now, the last term (l) or the nth term is given

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

Therefore, the sum of the 15 terms of the given A.P. is.

(iv) Here, we are given an A.P. whose nth term is given by the following expression, . We need to find the sum of first 15 terms.

So, here we can find the sum of the n terms of the given A.P., using the formula,

Where, a = the first term

l = the last term

So, for the given A.P,

The first term (a) will be calculated using in the given equation for nth term of A.P.

Now, the last term (l) or the nth term is given

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

Therefore, the sum of the 15 terms of the given A.P. is.

#### Question 6:

Find the sum of first 20 terms of the sequence whose nth term is an = An + B.

Here, we are given an A.P. whose nth term is given by the following expression. We need to find the sum of first 20 terms.

So, here we can find the sum of the n terms of the given A.P., using the formula,

Where, a = the first term

l = the last term

So, for the given A.P,

The first term (a) will be calculated using in the given equation for nth term of A.P.

Now, the last term (l) or the nth term is given

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

Therefore, the sum of the first 20 terms of the given A.P. is.

#### Question 7:

Find the sum of the first 25 terms of an A.P. whose nth term is given by an = 2 − 3n.

Here, we are given an A.P. whose nth term is given by the following expression, . We need to find the sum of first 25 terms.

So, here we can find the sum of the n terms of the given A.P., using the formula,

Where, a = the first term

l = the last term

So, for the given A.P,

The first term (a) will be calculated using in the given equation for nth term of A.P.

Now, the last term (l) or the nth term is given

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

Therefore, the sum of the 25 terms of the given A.P. is.

#### Question 8:

Find the sum of the first 25 terms of an A.P. whose nth term is given by an = 7 − 3n.

Here, we are given an A.P. whose nth term is given by the following expression, . We need to find the sum of first 25 terms.

So, here we can find the sum of the n terms of the given A.P., using the formula,

Where, a = the first term

l = the last term

So, for the given A.P,

The first term (a) will be calculated using in the given equation for nth term of A.P.

Now, the last term (l) or the nth term is given

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

Therefore, the sum of the 25 terms of the given A.P. is.

#### Question 9:

If the sum of a certain number of terms starting from first term of an A.P. is 25, 22, 19, ..., is 116. Find the last term.

In the given problem, we have the sum of the certain number of terms of an A.P. and we need to find the last term for that A.P.

So here, let us first find the number of terms whose sum is 116. For that, we will use the formula,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So for the given A.P

The first term (a) = 25

The sum of n terms

Common difference of the A.P. (d) =

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

So, we get the following quadratic equation,

On solving by splitting the middle term, we get,

Further,

Also,

Now, since n cannot be a fraction, so the number of terms is 8.

So, the term is a8

Therefore, the last term of the given A.P. such that the sum of the terms is 116 is.

#### Question 10:

(i) How many terms of the sequence 18, 16, 14, ... should be taken so that their sum is zero?

(ii) How many terms are there in the A.P. whose first and fifth terms are −14 and 2 respectively and the sum of the terms is 40?

(iii) How many terms of the A.P. 9, 17, 25,... must be taken so that their sum is 636?

(iv) How many terms of the A.P. 63, 60, 57, ... must be taken so that their sum is 693?

(v) How many terms of the A.P 27, 24, 21... should be taken so that their sum is zero?

(vi) How many terms of the A.P 45, 39, 33... must be taken so that their sum is 180?

In the given problem, we have the sum of the certain number of terms of an A.P. and we need to find the number of terms.

(i) A.P. is

So here, let us find the number of terms whose sum is 0. For that, we will use the formula,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

= number of terms

The first term (a) = 18

The sum of n terms (Sn) = 0

Common difference of the A.P. (d) =

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

Further,

Or,

Since, the number of terms cannot be zero, the number of terms (n) is.

(ii) Here, let us take the common difference as d.

So, we are given,

First term (a1) = −14

Fifth term (a5) = 2

Sum of terms (sn) = 40

Now,

Further, let us find the number of terms whose sum is 40. For that, we will use the formula,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

= number of terms

The first term (a1) = −14

The sum of n terms (Sn) = 40

Common difference of the A.P. (d) = 4

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

So, we get the following quadratic equation,

On solving by splitting the middle term, we get,

Further,

Or,

Since the number of terms cannot be negative. Therefore, the number of terms (n) is

(iii) A.P. is

So here, let us find the number of terms whose sum is 636. For that, we will use the formula,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

= number of terms

The first term (a) = 9

The sum of n terms (Sn) = 636

Common difference of the A.P. (d) =

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

So, we get the following quadratic equation,

On solving by splitting the middle term, we get,

Further,

Or,

Since, the number of terms cannot be a fraction, the number of terms (n) is.

(iv) A.P. is

So here, let us find the number of terms whose sum is 693. For that, we will use the formula,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

= number of terms

The first term (a) = 63

The sum of n terms (Sn) = 693

Common difference of the A.P. (d) =

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

So, we get the following quadratic equation,

On solving by splitting the middle term, we get,

Further,

Or,

Here, 22nd term will be

So, the sum of 22 as well as 21 terms is 693. Therefore, the number of terms (n) is.

(v)
The given AP is 27, 24, 21, ...
First term of the AP = 27
Common difference = 24 − 27 = −3
Let the sum of the first x terms of the AP be 0.
Sum of first x terms = $\frac{x}{2}\left[2×27+\left(x-1\right)\left(-3\right)\right]=0$
$⇒\frac{x}{2}\left[54+\left(-3x+3\right)\right]=0\phantom{\rule{0ex}{0ex}}⇒x\left(54-3x+3\right)=0\phantom{\rule{0ex}{0ex}}⇒x\left(57-3x\right)=0$
Now, either x = 0 or 57 − 3x = 0.
Since the number of terms cannot be 0, $x\ne 0$.
∴ 57 − 3x = 0
⇒ 57 = 3x
⇒ x = 19

Thus, the sum of the first 19 terms of the AP is 0.

(v) 27, 24, 21...

Hence, 19  terms of the A.P 27, 24, 21... should be taken so that their sum is zero.

#### Question 11:

Find the sum of the first

(i) 11 terms of the A.P.2, 6, 10. 14

(ii) 13 terms of the A.P. −6, 0, 6, 12, ...

(iii) 51 terms of the A.P. : whose second term is 2 and fourth term is 8.

In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of n terms of an A.P.,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

(i) To 11 terms.

Common difference of the A.P. (d) =

= 6 − 2
= 4

Number of terms (n) = 11

First term for the given A.P. (a) = 2

So, using the formula we get,

Therefore, the sum of first 11 terms for the given A.P. is.

(ii) To 13 terms.

Common difference of the A.P. (d) =

= 0 − (−6)
= 6

Number of terms (n) = 13

First term for the given A.P. (a) = −6

So, using the formula we get,

Therefore, the sum of first 13 terms for the given A.P. is.

(iii) 51 terms of an A.P whose and

Now,

Also,

Subtracting (1) from (2), we get

Further substituting in (1), we get

Number of terms (n) = 51

First term for the given A.P. (a) = −1

So, using the formula we get,

Therefore, the sum of first 51 terms for the given A.P. is.

#### Question 12:

Find the sum of

(i) the first 15 multiples of 8

(ii) the first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.

(iii) all 3 − digit natural numbers which are divisible by 13.

(iv) all 3 − digit natural numbers, which are multiples of 11.

(v) all 2 − digit natural numbers divisible by 4.

(vi) first 8 multiples of 3.

In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of n terms of an A.P.,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

(i) First 15 multiples of 8.

So, we know that the first multiple of 8 is 8 and the last multiple of 8 is 120.

Also, all these terms will form an A.P. with the common difference of 8.

So here,

First term (a) = 8

Number of terms (n) = 15

Common difference (d) = 8

Now, using the formula for the sum of n terms, we get

Therefore, the sum of the first 15 multiples of 8 is

(ii) (a) First 40 positive integers divisible by 3

So, we know that the first multiple of 3 is 3 and the last multiple of 3 is 120.

Also, all these terms will form an A.P. with the common difference of 3.

So here,

First term (a) = 3

Number of terms (n) = 40

Common difference (d) = 3

Now, using the formula for the sum of n terms, we get

Therefore, the sum of first 40 multiples of 3 is

(b) First 40 positive integers divisible by 5

So, we know that the first multiple of 5 is 5 and the last multiple of 5 is 200.

Also, all these terms will form an A.P. with the common difference of 5.

So here,

First term (a) = 5

Number of terms (n) = 40

Common difference (d) = 5

Now, using the formula for the sum of n terms, we get

Therefore, the sum of first 40 multiples of 3 is

(c) First 40 positive integers divisible by 6

So, we know that the first multiple of 6 is 6 and the last multiple of 6 is 240.

Also, all these terms will form an A.P. with the common difference of 6.

So here,

First term (a) = 6

Number of terms (n) = 40

Common difference (d) = 6

Now, using the formula for the sum of n terms, we get

Therefore, the sum of first 40 multiples of 3 is

(iii) All 3 digit natural number which are divisible by 13

So, we know that the first 3 digit multiple of 13 is 104 and the last 3 digit multiple of 13 is 988.

Also, all these terms will form an A.P. with the common difference of 13.

So here,

First term (a) = 104

Last term (l) = 988

Common difference (d) = 13

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of n terms, we get

On further simplification, we get,

Therefore, the sum of all the 3 digit multiples of 13 is.
(iv) all 3-digit natural numbers, which are multiples of 11.

We know that the first 3 digit number multiple of 11 will be 110.
Last 3 digit number multiple of 11 will be 990.
So here,

First term (a) = 110

Last term (l) = 990

Common difference (d) = 11

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

So, for the last term,

$990=110+\left(n-1\right)11\phantom{\rule{0ex}{0ex}}990=110+11n-11\phantom{\rule{0ex}{0ex}}990=99+11n\phantom{\rule{0ex}{0ex}}891=11n\phantom{\rule{0ex}{0ex}}81=n$

Now, using the formula for the sum of n terms, we get

${S}_{n}=\frac{81}{2}\left[2\left(110\right)+\left(81-1\right)11\right]\phantom{\rule{0ex}{0ex}}{S}_{n}=\frac{81}{2}\left[220+80×11\right]\phantom{\rule{0ex}{0ex}}{S}_{n}=\frac{81}{2}×1100\phantom{\rule{0ex}{0ex}}{S}_{n}=81×550\phantom{\rule{0ex}{0ex}}{S}_{n}=44550$

Therefore, the sum of all the 3 digit multiples of 11 is 44550.

(v) 2-digit no. divisible by 4 are 12,16,20,........,96
We can see it forms an AP as the common difference is 4 and the first term is 4.
To find no. of terms n,
we know that

$96=12+\left(n-1\right)4\phantom{\rule{0ex}{0ex}}84=\left(n-1\right)4\phantom{\rule{0ex}{0ex}}21=n-1\phantom{\rule{0ex}{0ex}}22=n$
Now,

First term (a) = 12

Number of terms (n) = 22

Common difference (d) =4

Now, using the formula for the sum of n terms, we get
${S}_{22}=\frac{22}{2}\left\{2\left(12\right)+\left(22-1\right)4\right\}\phantom{\rule{0ex}{0ex}}{S}_{22}=11\left\{24+84\right\}\phantom{\rule{0ex}{0ex}}{S}_{22}=1188$

Hence, the sum of 22 terms is 1188 which are divisible by 4.

(vi)
First 8 multiples of 3 are { 3, 6, 9...,24}

We can observe they are in AP with first term (a) = 3 and last term (l) = 24 and number of terms are 8.

Hence, the sum of the first 8 multiples of 3 is 108.

#### Question 13:

Find the sum:

(i) 2 + 4 + 6 ... + 200

(ii) 3 + 11 + 19 + ... + 803

(iii) (−5) + (−8)+ (−11) + ... + (−230)

(iv) 1 + 3 + 5 + 7 + ... + 199

(v)

(vi) 34 + 32 + 30 + ... + 10

(vii) 25 + 28 + 31 + ... + 100

(viii)

In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of n terms of an A.P.,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

(i)

Common difference of the A.P. (d) =

So here,

First term (a) = 2

Last term (l) = 200

Common difference (d) = 2

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of n terms, we get

On further simplification, we get,

Therefore, the sum of the A.P is

(ii)

Common difference of the A.P. (d) =

So here,

First term (a) = 3

Last term (l) = 803

Common difference (d) = 8

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of n terms, we get

Therefore, the sum of the A.P is

(iii)

Common difference of the A.P. (d) =

So here,

First term (a) = −5

Last term (l) = −230

Common difference (d) = −3

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

So, for the last term,

Now, using the formula for the sum of n terms, we get

Therefore, the sum of the A.P is

(iv)

Common difference of the A.P. (d) =

So here,

First term (a) = 1

Last term (l) = 199

Common difference (d) = 2

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

So, for the last term,

Now, using the formula for the sum of n terms, we get

On further simplification, we get,

Therefore, the sum of the A.P is

(v)

Common difference of the A.P is

(d) =

So here,

First term (a) = 7

Last term (l) = 84

Common difference (d) =

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

So, for the last term,

Further solving for n,

Now, using the formula for the sum of n terms, we get

On further simplification, we get,

Therefore, the sum of the A.P is

(vi)

Common difference of the A.P. (d) =

So here,

First term (a) = 34

Last term (l) = 10

Common difference (d) = −2

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

So, for the last term,

Further solving for n,

Now, using the formula for the sum of n terms, we get

On further simplification, we get,

Therefore, the sum of the A.P is

(vii)

Common difference of the A.P. (d) =

So here,

First term (a) = 25

Last term (l) = 100

Common difference (d) = 3

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

So, for the last term,

Further solving for n,

Now, using the formula for the sum of n terms, we get

On further simplification, we get,

Therefore, the sum of the A.P is.
(viii) $18+15\frac{1}{2}+13+...+\left(-49\frac{1}{2}\right)$

Common difference of the A.P. (d) =

$=15\frac{1}{2}-18\phantom{\rule{0ex}{0ex}}=\frac{31}{2}-18\phantom{\rule{0ex}{0ex}}=\frac{31-36}{2}\phantom{\rule{0ex}{0ex}}=\frac{-5}{2}$

So here,

First term (a) = 18

Last term (l) = $-49\frac{1}{2}=\frac{-99}{2}$

Common difference (d) = $\frac{-5}{2}$

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

So, for the last term,

$\frac{-99}{2}=18+\left(n-1\right)\frac{-5}{2}\phantom{\rule{0ex}{0ex}}\frac{-99}{2}=18+\left(\frac{-5}{2}\right)n+\frac{5}{2}\phantom{\rule{0ex}{0ex}}\frac{5}{2}n=18+\frac{5}{2}+\frac{99}{2}\phantom{\rule{0ex}{0ex}}\frac{5}{2}n=18+\frac{104}{2}\phantom{\rule{0ex}{0ex}}n=28$

Now, using the formula for the sum of n terms, we get

${S}_{n}=\frac{28}{2}\left[2×18+\left(28-1\right)\left(\frac{-5}{2}\right)\right]\phantom{\rule{0ex}{0ex}}{S}_{n}=14\left[36+27\left(\frac{-5}{2}\right)\right]\phantom{\rule{0ex}{0ex}}{S}_{n}=-441\phantom{\rule{0ex}{0ex}}$

Therefore, the sum of the A.P is ${S}_{n}=-441$.

#### Question 14:

The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

In the given problem, we have the first and the last term of an A.P. along with the common difference of the A.P. Here, we need to find the number of terms of the A.P. and the sum of all the terms.

Here,

The first term of the A.P (a) = 17

The last term of the A.P (l) = 350

The common difference of the A.P. = 9

Let the number of terms be n.

So, as we know that,

We get,

Further solving this,

Using the above values in the formula,

Therefore, the number of terms is and the sum.

#### Question 15:

The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference and the sum of first 20 terms.

In the given problem, let us take the first term as a and the common difference as d.

Here, we are given that,

So, using (1) in (2), we get,

Also, we know,

For the 3th term (n = 3),

Similarly, for the 7th term (n = 7),

Subtracting (4) from (5), we get,

Now, to find a, we substitute the value of d in (4),

So, for the given A.P

So, to find the sum of first 20 terms of this A.P., we use the following formula for the sum of n terms of an A.P.,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 20, we get,

Therefore, the sum of first 20 terms for the given A.P. is .

#### Question 16:

The first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the common difference.

In the given problem, we have the first and the last term of an A.P. along with the sum of all the terms of A.P. Here, we need to find the common difference of the A.P.

Here,

The first term of the A.P (a) = 2

The last term of the A.P (l) = 50

Sum of all the terms

Let the common difference of the A.P. be d.

So, let us first find the number of the terms (n) using the formula,

Now, to find the common difference of the A.P. we use the following formula,

We get,

Further, solving for d,

Therefore the common difference of the A.P.

#### Question 17:

If 12th term of an A.P. is −13 and the sum of the first four terms is 24, what is the sum of first 10 terms.

In the given problem, we need to find the sum of first 10 terms of an A.P. Let us take the first term a and the common difference as d

Here, we are given that,

Also, we know,

For the 12th term (n = 12),

So, as we know the formula for the sum of n terms of an A.P. is given by,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 4, we get,

Subtracting (1) from (2), we get,

On further simplifying for d, we get,

Now, to find a, we substitute the value of d in (1),

Now, using the formula for the sum of n terms of an A.P. for n = 10, we get,

Therefore, the sum of first 10 terms for the given A.P. is .

#### Question 18:

Find the sum of n terms of the series $\left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+..........$

Let the given series be S = $\left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+..........$
$=\left[4+4+4+...\right]-\left[\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+...\right]\phantom{\rule{0ex}{0ex}}=4\left[1+1+1+...\right]-\frac{1}{n}\left[1+2+3+...\right]\phantom{\rule{0ex}{0ex}}={S}_{1}-{S}_{2}$

Hence, the sum of n terms of the series is $\frac{7n-1}{2}$.

#### Question 19:

In an A.P., if the first term is 22, the common difference is −4 and the sum to n terms is 64,  find n.

In the given problem, we need to find the number of terms of an A.P. Let us take the number of terms as n.

Here, we are given that,

So, as we know the formula for the sum of n terms of an A.P. is given by,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula we get,

Further rearranging the terms, we get a quadratic equation,

On taking 4 common, we get,

Further, on solving the equation for n by splitting the middle term, we get,

So, we get,

Also,

Therefore,

#### Question 20:

In an A.P., if the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 terms?

In the given problem, let us take the first term as a and the common difference d

Here, we are given that,

Also, we know,

For the 5th term (n = 5),

Similarly, for the 12th term (n = 12),

Subtracting (3) from (4), we get,

Now, to find a, we substitute the value of d in (4),

So, for the given A.P

So, to find the sum of first 20 terms of this A.P., we use the following formula for the sum of n terms of an A.P.,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 20, we get,

Therefore, the sum of first 20 terms for the given A.P. is.

#### Question 21:

Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.

In the given problem, let us take the first term as a and the common difference as d.

Here, we are given that,

Also, we know,

For the 2nd term (n = 2),

Similarly, for the 3rd term (n = 3),

Subtracting (3) from (4), we get,

Now, to find a, we substitute the value of d in (4),

So, for the given A.P

So, to find the sum of first 51 terms of this A.P., we use the following formula for the sum of n terms of an A.P.,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 51, we get,

Therefore, the sum of first 51 terms for the given A.P. is.

#### Question 22:

(i) If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of n terms.

(ii) If the sum of first four terms of an A.P. is 40 and that of first 14 terms is 280. Find the sum of its first n terms.

(i)

In the given problem, we need to find the sum of n terms of an A.P. Let us take the first term as a and the common difference as d.

Here, we are given that,

So, as we know the formula for the sum of n terms of an A.P. is given by,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

= number of terms

So, using the formula for n = 7, we get,

Further simplifying for a, we get,

Also, using the formula for n = 17, we get,

Further simplifying for a, we get,

Subtracting (3) from (4), we get,

Now, to find a, we substitute the value of d in (3),

Now, using the formula for the sum of n terms of an A.P., we get,

Therefore, the sum of first n terms for the given A.P. is.

(ii) Given:

Hence,  the sum of its first n terms is n2 + 6n.

#### Question 23:

The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

In the given problem, we have the first and the last term of an A.P. along with the sum of all the terms of A.P. Here, we need to find the number of terms and the common difference of the A.P.

Here,

The first term of the A.P (a) = 5

The last term of the A.P (l) = 45

Sum of all the terms

Let the common difference of the A.P. be d.

So, let us first find the number of the terms (n) using the formula,

Now, to find the common difference of the A.P. we use the following formula,

We get,

Further, solving for d,

Therefore, the number of terms is and the common difference of the A.P is.

#### Question 24:

In an A.P. the first term is 8, nth term is 33 and the sum to first n terms is 123. Find n and d, the common differences.

In the given problem, we have the first and the nth term of an A.P. along with the sum of the n terms of A.P. Here, we need to find the number of terms and the common difference of the A.P.

Here,

The first term of the A.P (a) = 8

The nth term of the A.P (l) = 33

Sum of all the terms

Let the common difference of the A.P. be d.

So, let us first find the number of the terms (n) using the formula,

Now, to find the common difference of the A.P. we use the following formula,

We get,

Further, solving for d,

Therefore, the number of terms is and the common difference of the A.P..

#### Question 25:

In an A.P., the first term is 22, nth term is −11 and the sum to first n terms is 66. Find n and d, the common difference

In the given problem, we have the first and the nth term of an A.P. along with the sum of the n terms of A.P. Here, we need to find the number of terms and the common difference of the A.P.

Here,

The first term of the A.P (a) = 22

The nth term of the A.P (l) = −11

Sum of all the terms

Let the common difference of the A.P. be d.

So, let us first find the number of the terms (n) using the formula,

Further, solving for n

Now, to find the common difference of the A.P. we use the following formula,

We get,

Further, solving for d,

Therefore, the number of terms is and the common difference of the A.P..

#### Question 26:

​The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find its common difference.

Let a be the first term and d be the common difference.

We know that, sum of first n terms = Sn $\frac{n}{2}$[2a + (n − 1)d]

Also, nth term = an = a + (n − 1)d

According to the question,

a = 7, an = 49 and Sn = 420

Now,
an = a + (n − 1)d
49 = 7 + (n − 1)d
​⇒ 42 = nd − d
​⇒
nd − = 42                     ....(1)

Also,
Sn $\frac{n}{2}$[2 × 7 + (n − 1)d]
⇒ 420 = $\frac{n}{2}$[14 + nd − d]
⇒ 840 = n[14 + 42]               [From (1)]
⇒ 56n = 840
⇒ n = 15                              ....(2)

On substituting (2) in (1), we get

nd − = 42
⇒ (15 − 1)= 42
⇒ 14= 42
⇒ = 3

Thus, common difference of the given A.P. is 3.

#### Question 27:

The first and the last terms of an A.P. are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference.

Let a be the first term and d be the common difference.

We know that, sum of first n terms = Sn $\frac{n}{2}$[2a + (n − 1)d]

Also, nth term = an = a + (n − 1)d

According to the question,

a = 5, an = 45 and Sn = 400

Now,
an = a + (n − 1)d
45 = 5 + (n − 1)d
​⇒ 40 = nd − d
​⇒
nd − = 40                     ....(1)

Also,
Sn $\frac{n}{2}$[2 × 5 + (n − 1)d]
⇒ 400 = $\frac{n}{2}$[10 + nd − d]
⇒ 800 = n[10 + 40]               [From (1)]
⇒ 50n = 800
n = 16                              ....(2)

On substituting (2) in (1), we get

nd − = 40
⇒ (16 − 1)= 40
⇒ 15= 40
$\frac{8}{3}$

Thus, common difference of the given A.P. is $\frac{8}{3}$.

#### Question 28:

The sum of first 9 terms of an A.P. is 162. The ratio of its 6th term to its 13th term is 1 : 2. Find the first and 15th term of the A.P.

Disclaimer: There is a misprint in the question, 'the sum of 9 terms' should be witten instead of 'the sum of q terms'.

Let a be the first term and d be the common difference.

We know that, sum of first n terms = Sn $\frac{n}{2}$[2a + (n − 1)d]

Also, nth term = an = a + (n − 1)d

According to the question,

Sq = 162 and $\frac{{a}_{6}}{{a}_{13}}=\frac{1}{2}$

Now,

Also,
S$\frac{9}{2}$[2a + (9 − 1)d]
⇒ 162 = $\frac{9}{2}$[2(2d) + 8d]           [From (1)]
⇒ 18 = $\frac{1}{2}$[12d]
⇒ 18 = 6d
⇒ d = 3
⇒ a = 2 × 3                              [From (1)]
⇒ a = 6

Thus, the first term of the A.P. is 6.

Now,

a15 = 6 + (15 − 1)3
= 6 + 42
= 48

Thus, 15th term of the A.P. is 48.

#### Question 29:

If the 10th term of an A.P. is 21 and the sum of its first 10 terms is 120, find its nth term.

Let a be the first term and d be the common difference.

We know that, sum of first n terms = Sn $\frac{n}{2}$[2a + (n − 1)d]
and nth term = an = a + (n − 1)d

Now,
S10 $\frac{10}{2}$[2a + (10 − 1)d]
⇒ 120 = 5(2a + 9d)
⇒ 24 = 2a + 9d
⇒ 2a + 9d = 24                    ....(1)

Also,
a10 = a + (10 − 1)d
⇒ 21 = a + 9d
⇒ 2a + 18= 42                  ....(2)

Subtracting (1) from (2), we get
18d − 9d = 42 − 24
⇒ 9d = 18
⇒ d = 2
⇒ 2a = 24 − 9d             [From (1)]
⇒ 2a = 24 − 9 × 2
⇒ 2a = 24 − 18
⇒ 2a = 6
a = 3

Also,
an = a + (n − 1)d
=
3 + (− 1)2
= 3 + 2− 2
= 1 + 2n

Thus, nth term of this A.P. is 1 + 2n.

#### Question 30:

The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P.

Let a be the first term and d be the common difference.

We know that, sum of first n terms = Sn $\frac{n}{2}$[2a + (n − 1)d]

It is given that sum of the first 7 terms of an A.P. is 63.
And sum of next 7 terms is 161.
∴ Sum of first 14 terms = Sum of first 7 terms + sum of next 7 terms
= 63 + 161 = 224

Now,
S$\frac{7}{2}$[2a + (7 − 1)d]
⇒ 63 = $\frac{7}{2}$(2a + 6d)
⇒ 18 = 2a + 6d
⇒ 2a + 6d = 18         ....(1)

Also,
S14 $\frac{14}{2}$[2a + (14 − 1)d]
⇒ 224 = 7(2a + 13d)
⇒ 32 = 2a + 13d
⇒ 2a + 13d =  32      ....(2)

On subtracting (1) from (2), we get
13d − 6d = 32 − 18
⇒ 7d = 14
d = 2
⇒ 2a = 18 − 6d         [From (1)]
⇒ 2a = 18 − 6 × 2
⇒ 2a = 18 − 12
⇒ 2a = 6
a = 3

Also, nth term = an = a + (n − 1)d
⇒ a28 = 3 + (28 − 1)2
= 3 + 27 × 2
= 57

Thus, 28th term of this A.P. is 57.

#### Question 31:

The sum of first seven terms of an A.P. is 182. If its 4th and the 17th terms are in the ratio 1 : 5, find the A.P.

Let a be the first term and d be the common difference.

We know that, sum of first n terms = Sn $\frac{n}{2}$[2a + (n − 1)d]

According to the question,

Also,

On substituting (2) in (1), we get

Thus, the A.P. is 2, 10, 18, 26, ..... .

#### Question 32:

The nth term of an A.P is given by (−4n + 15), Find the sum of first 20 terms of this A.P.

#### Question 33:

In an A.P., the sum of first ten terms is −150 and the sum of its next ten terms is −550. Find the A.P.

Here, we are given  and sum of the next ten terms is −550.

Let us take the first term of the A.P. as a and the common difference as d.

So, let us first find a10. For the sum of first 10 terms of this A.P,

First term = a

Last term = a10

So, we know,

For the 10th term (n = 10),

So, here we can find the sum of the n terms of the given A.P., using the formula,

Where, a = the first term

l = the last term

So, for the given A.P,

Similarly, for the sum of next 10 terms (S10),

First term = a11

Last term = a20

For the 11th term (n = 11),

For the 20th term (n = 20),

So, for the given A.P,

Now, subtracting (1) from (2),

Substituting the value of in (1)

So, the A.P. is  with .

#### Question 34:

Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25th term.

First term, a = 10

Sum of first 14 terms, ${S}_{14}=1505$

$⇒\frac{14}{2}\left[2×10+\left(14-1\right)d\right]=1505\phantom{\rule{0ex}{0ex}}⇒7×\left(20-13d\right)=1505\phantom{\rule{0ex}{0ex}}⇒20-13d=\frac{1505}{7}=215\phantom{\rule{0ex}{0ex}}⇒13d=-195\phantom{\rule{0ex}{0ex}}⇒d=-15$

Now,
${a}_{25}=10+24\left(-15\right)=-350$

#### Question 35:

In an A.P., the first term is 2, the last term is 29 and the sum of the terms is 155. Find the common difference of the A.P.

In the given problem, we have the first and the last term of an A.P. along with the sum of all the terms of A.P. Here, we need to find the common difference of the A.P.

Here,

The first term of the A.P (a) = 2

The last term of the A.P (l) = 29

Sum of all the terms (Sn) = 155

Let the common difference of the A.P. be d.

So, let us first find the number of the terms (n) using the formula,

Now, to find the common difference of the A.P. we use the following formula,

We get,

Therefore, the common difference of the A.P. is.

#### Question 36:

The first and the last term of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

In the given problem, we have the first and the last term of an A.P. along with the common difference of the A.P. Here, we need to find the number of terms of the A.P. and the sum of all the terms.

Here,

The first term of the A.P (a) = 17

The last term of the A.P (l) = 350

The common difference of the A.P. = 9

Let the number of terms be n.

So, as we know that,

We get,

Further solving this,

Using the above values in the formula,

Therefore, the number of terms is and the sum.

#### Question 37:

Find the number of terms of the A.P. −12, −9, −6, ..., 21. If 1 is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained.

First term, ${a}_{1}=-12$
Common difference, $d={a}_{2}-{a}_{1}=-9-\left(-12\right)=3$
${a}_{n}=21\phantom{\rule{0ex}{0ex}}⇒a+\left(n-1\right)d=21\phantom{\rule{0ex}{0ex}}⇒-12+\left(n-1\right)×3=21\phantom{\rule{0ex}{0ex}}⇒3n=36\phantom{\rule{0ex}{0ex}}⇒n=12$

Therefore, number of terms in the given A.P. is 12.

Now, when 1 is added to each of the 12 terms, the sum will increase by 12.

So, the sum of all terms of the A.P. thus obtained

$={S}_{12}+12\phantom{\rule{0ex}{0ex}}=\frac{12}{2}\left[2\left(-12\right)+11\left(3\right)\right]+12\phantom{\rule{0ex}{0ex}}=6×\left(9\right)+12\phantom{\rule{0ex}{0ex}}=66\phantom{\rule{0ex}{0ex}}$

#### Question 38:

The sum of the first n terms of an A.P. is 3n2 + 6n. Find the nth term of this A.P.

Let a be the first term and d be the common difference.

We know that, sum of first n terms = Sn $\frac{n}{2}$[2a + (n − 1)d]

It is given that sum of the first n terms of an A.P. is 3n2 + 6n.

∴ First term = a =  S= 3(1)2 + 6(1) = 9.

Sum of first two terms = S= 3(2)2 + 6(2) = 24.

∴ Second term = S2 − S1 = 24 − 9 = 15.

∴ Common difference = d = Second term − First term
= 15 − 9 = 6

Also, nth term = an = a + (n − 1)d
⇒ an = 9 + (n − 1)6
⇒ an = 9 + 6n − 6
⇒ an = 3 + 6n

Thus, nth term of this A.P. is 3 + 6n.

#### Question 39:

The sum of first n terms of an A.P. is 5n − n2. Find the nth term of this A.P.

Let a be the first term and d be the common difference.

We know that, sum of first n terms = Sn $\frac{n}{2}$[2a + (n − 1)d]

It is given that sum of the first n terms of an A.P. is 5n − n2.

∴ First term = =  S= 5(1) − (1)2 = 4.

Sum of first two terms = S= 5(2) − (2)2 = 6.

∴ Second term = S2 − S1 = 6 − 4 = 2.

∴ Common difference = d = Second term − First term
= 2 − 4 = −2

Also, nth term = an = a + (n − 1)d
⇒ an = 4 + (n − 1)(−2)
⇒ an = 4 − 2n + 2
⇒ an = 6 − 2n

Thus, nth term of this A.P. is 6 − 2n.

#### Question 40:

The sum of the first n terms of an A.P. is 4n2 + 2n. Find the nth term of this A.P.

Let a be the first term and d be the common difference.

We know that, sum of first n terms = Sn $\frac{n}{2}$[2a + (n − 1)d]

It is given that sum of the first n terms of an A.P. is 4n2 + 2n.

∴ First term = =  S= 4(1)2 + 2(1) = 6.

Sum of first two terms = S= 4(2)2 + 2(2) = 20.

∴ Second term = S2 − S1 = 20 − 6 = 14.

∴ Common difference = d = Second term − First term
= 14 − 6 = 8

Also, nth term = an = a + (n − 1)d
⇒ an = 6 + (n − 1)(8)
⇒ an = 6 + 8n − 8
⇒ an = 8− 2

Thus, nth term of this A.P. is 8− 2.

#### Question 41:

The sum of first n terms of an A.P. is 3n2 + 4n. Find the 25th term of this A.P.

${S}_{n}=3{n}^{2}+4n$

We know

${a}_{n}={S}_{n}-{S}_{n-1}\phantom{\rule{0ex}{0ex}}\therefore {a}_{n}=3{n}^{2}+4n-3{\left(n-1\right)}^{2}-4\left(n-1\right)\phantom{\rule{0ex}{0ex}}⇒{a}_{n}=6n+1$

${a}_{25}=6\left(25\right)+1=151$

#### Question 42:

The sum of first n terms of an A.P is 5n2 + 3n. If its mth terms is 168, find the value of m. Also, find the 20th term of this A.P.

Now,

${a}_{m}=168\phantom{\rule{0ex}{0ex}}⇒10m-2=168\phantom{\rule{0ex}{0ex}}⇒10m=170\phantom{\rule{0ex}{0ex}}⇒m=17$

${a}_{20}=10\left(20\right)-2=198$

#### Question 43:

The sum of first q terms of an A.P. is 63q − 3q2. If its pth term is −60,  find the value of p. Also, find the 11th term of this A.P.

#### Question 44:

The sum of first m terms of an A.P. is 4m2m. If its nth term is 107. find the value of n. Also, find the 21st term of this A.P.

#### Question 45:

If the sum of the first n terms of an A.P is 4n − n2, What is the first term? What is the sum of first two terms? What is the second term? Similarly, find the third, the tenth and the nth terms.

In the given problem, the sum of n terms of an A.P. is given by the expression,

So here, we can find the first term by substituting,

Similarly, the sum of first two terms can be given by,

Now, as we know,

So,

Now, using the same method we have to find the third, tenth and nth term of the A.P.

So, for the third term,

Also, for the tenth term,

So, for the nth term,

Therefore, .

#### Question 46:

If the sum of first n terms of an A.P. is $\frac{1}{2}$(3n2 + 7n), then find its nth term. Hence write its 20th term.

Let a be the first term and d be the common difference.

We know that, sum of first n terms = Sn $\frac{n}{2}$[2a + (n − 1)d]

It is given that sum of the first n terms of an A.P. is $\frac{1}{2}$(3n2 + 7n).

∴ First term = =  S$\frac{1}{2}$[3(1)2 + 7(1)] = 5.

Sum of first two terms = S$\frac{1}{2}$[3(2)2 + 7(2)] = 13.

∴ Second term = S2 − S1 = 13 − 5 = 8.

∴ Common difference = d = Second term − First term
= 8 − 5 = 3

Also, nth term = an = a + (n − 1)d
⇒ an = 5 + (n − 1)(3)
⇒ an = 5 + 3n − 3
⇒ an = 3+ 2

Thus, nth term of this A.P. is 3+ 2.

Now,
a20 = a + (20 − 1)d
⇒ a20 = 5 + 19(3)
⇒ a20 = 5 + 57
⇒ a20 = 62

Thus, 20th term of this A.P is 62.

#### Question 47:

(i) In an A.P., the sum of first n terms is $\frac{3{n}^{2}}{2}+\frac{13}{n}n.$ Find its 25th term.

(ii) If the sum of first n terms of an A.P. is n2, then find its 10th term.

(i)

Here, the sum of first n terms is given by the expression,

We need to find the 25th term of the A.P.

So we know that the nthterm of an A.P. is given by,

So …… (1)

So, using the expression given for the sum of n terms, we find the sum of 25 terms (S25) and the sum of 24 terms (S24). We get,

Similarly,

Now, using the above values in (1),

Therefore, .

(ii) Given:

Hence, the 10th term is 19.

#### Question 48:

Find the sum of all natural numbers between 1 and 100, which are divisible by 3.

In this problem, we need to find the sum of all the multiples of 3 lying between 1 and 100.

So, we know that the first multiple of 3 after 1 is 3 and the last multiple of 3 before 100 is 99.

Also, all these terms will form an A.P. with the common difference of 3.

So here,

First term (a) = 3

Last term (l) = 99

Common difference (d) = 3

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of n terms,

We get,

On further simplification, we get,

Therefore, the sum of all the multiples of 3 lying between 1 and 100 is

#### Question 49:

Find the sum of first n odd natural numbers.

In this problem, we need to find the sum of first n odd natural numbers.

So, we know that the first odd natural number is 1. Also, all the odd terms will form an A.P. with the common difference of 2.

So here,

First term (a) = 1

Common difference (d) = 2

So, let us take the number of terms as n

Now, as we know,

So, for n terms,

Therefore, the sum of first n odd natural numbers is.

#### Question 50:

Find the sum of all odd numbers between (i) 0 and 50 (ii) 100 and 200.

(i) In this problem, we need to find the sum of all odd numbers lying between 0 and 50.

So, we know that the first odd number after 0 is 1 and the last odd number before 50 is 49.

Also, all these terms will form an A.P. with the common difference of 2.

So here,

First term (a) = 1

Last term (l) = 49

Common difference (d) = 2

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of n terms,

For n = 25, we get,

On further simplification, we get,

Therefore, the sum of all the odd numbers lying between 0 and 50 is.

(ii) In this problem, we need to find the sum of all odd numbers lying between 100 and 200.

So, we know that the first odd number after 0 is 101 and the last odd number before 200 is 199.

Also, all these terms will form an A.P. with the common difference of 2.

So here,

First term (a) = 101

Last term (l) = 199

Common difference (d) = 2

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of n terms,

For n = 50, we get,

Therefore, the sum of all the odd numbers lying between 100 and 200 is.

#### Question 51:

Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.

In this problem, we need to prove that the sum of all odd numbers lying between 1 and 1000 which are divisible by 3 is 83667.

So, we know that the first odd number after 1 which is divisible by 3 is 3, the next odd number divisible by 3 is 9 and the last odd number before 1000 is 999.

So, all these terms will form an A.P. 3, 9, 15, 21 … with the common difference of 6

So here,

First term (a) = 3

Last term (l) = 999

Common difference (d) = 6

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of n terms,

For n = 167, we get,

On further simplification, we get,

Therefore, the sum of all the odd numbers lying between 1 and 1000 is.

Hence proved

#### Question 52:

Find the sum of all integers between 84 and 719, which are multiples of 5.

In this problem, we need to find the sum of all the multiples of 5 lying between 84 and 719.

So, we know that the first multiple of 5 after 84 is 85 and the last multiple of 5 before 719 is 715.

Also, all these terms will form an A.P. with the common difference of 5.

So here,

First term (a) = 85

Last term (l) = 715

Common difference (d) = 5

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of n terms,

We get,

On further simplification, we get,

Therefore, the sum of all the multiples of 5 lying between 84 and 719 is.

#### Question 53:

Find the sum of all integers between 50 and 500, which are divisible by 7.

In this problem, we need to find the sum of all the multiples of 7 lying between 50 and 500.

So, we know that the first multiple of 7 after 50 is 56 and the last multiple of 7 before 500 is 497.

Also, all these terms will form an A.P. with the common difference of 7.

So here,

First term (a) = 56

Last term (l) = 497

Common difference (d) = 7

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of n terms,

For n = 64, we get,

Therefore, the sum of all the multiples of 7 lying between 50 and 500 is.

#### Question 54:

Find the sum of all even integers between 101 and 999.

In this problem, we need to find the sum of all the even numbers lying between 101 and 999.

So, we know that the first even number after 101 is 102 and the last even number before 999 is 998.

Also, all these terms will form an A.P. with the common difference of 2.

So here,

First term (a) = 102

Last term (l) = 998

Common difference (d) = 2

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of n terms,

For n = 64, we get,

On further simplification, we get,

Therefore, the sum of all the even numbers lying between 101 and 999 is.

#### Question 55:

Find the sum of all integers between 100 and 550, which are divisible by 9.

(i) In this problem, we need to find the sum of all the multiples of 9 lying between 100 and 550.

So, we know that the first multiple of 9 after 100 is 108 and the last multiple of 9 before 550 is 549.

Also, all these terms will form an A.P. with the common difference of 9.

So here,

First term (a) = 108

Last term (l) = 549

Common difference (d) = 9

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of n terms,

We get,

Therefore, the sum of all the multiples of 9 lying between 100 and 550 is

(ii)
In this problem, we need to find the sum of all the integers lying between 100 and 550 which are not multiples of 9.
So, we know that the sum of all the multiples of 9 lying between 100 and 550 is 16425.
The sum of all the integers lying between 100 and 550 which are not multiples of 9 is

[101 + 102 + 103 + .....+ 549] − 16425
$=\left[\left(1+2+.....+549\right)-\left(1+2+.....100\right)\right]-16425\phantom{\rule{0ex}{0ex}}=\left[\left(\frac{549×550}{2}\right)-\left(\frac{100×101}{2}\right)\right]-16425\phantom{\rule{0ex}{0ex}}=150975-5050-16425\phantom{\rule{0ex}{0ex}}=129500\phantom{\rule{0ex}{0ex}}$

(iii)
Integers between 1 and 500 which are multiples of both 2 and 5 are
10, 20, 30, ....., 490.
${a}_{n}=490\phantom{\rule{0ex}{0ex}}⇒10+\left(n-1\right)10=490\phantom{\rule{0ex}{0ex}}⇒\left(n-1\right)10=480\phantom{\rule{0ex}{0ex}}⇒n=49\phantom{\rule{0ex}{0ex}}Now,\phantom{\rule{0ex}{0ex}}{S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}{S}_{49}=\frac{49}{2}\left[20+\left(48\right)10\right]\phantom{\rule{0ex}{0ex}}{S}_{49}=\frac{49}{2}\left[500\right]\phantom{\rule{0ex}{0ex}}{S}_{49}=49×250=12250$

(iv)
Integers from 1 and 500 which are multiples of both 2 and 5 are
10, 20, 30, ....., 500.
${a}_{n}=500\phantom{\rule{0ex}{0ex}}⇒10+\left(n-1\right)10=500\phantom{\rule{0ex}{0ex}}⇒\left(n-1\right)10=490\phantom{\rule{0ex}{0ex}}⇒n=50\phantom{\rule{0ex}{0ex}}Now,\phantom{\rule{0ex}{0ex}}{S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}{S}_{50}=\frac{50}{2}\left[20+\left(49\right)10\right]\phantom{\rule{0ex}{0ex}}=25\left[510\right]\phantom{\rule{0ex}{0ex}}=12750$

(v)

Integers from 1 and 500 which are multiples of 2 or 5 are
Multiples of 2 are 2, 4, 6, .....500. Therefore sum of all the multiples of 2 from 1 and 500 is
${a}_{n}=500\phantom{\rule{0ex}{0ex}}⇒2+\left(n-1\right)2=500\phantom{\rule{0ex}{0ex}}⇒n=250\phantom{\rule{0ex}{0ex}}{S}_{250}=\frac{250}{2}\left(4+249×2\right)=125×502=62750\phantom{\rule{0ex}{0ex}}$
and
Multiples of 5 are 5, 10, 15, ....., 500. Therefore sum of all the multiples of 5 from 1 and 500 is
${a}_{n}=500\phantom{\rule{0ex}{0ex}}⇒5+\left(n-1\right)5=500\phantom{\rule{0ex}{0ex}}⇒n=100\phantom{\rule{0ex}{0ex}}{S}_{100}=\frac{100}{2}\left(10+99×5\right)=50×505=25250\phantom{\rule{0ex}{0ex}}$
Multiples of both 2 and 5 are 10, 20, 30, ....., 500. Therefore sum of all the multiples of 2 and 5 from 1 and 500 is
${a}_{n}=500\phantom{\rule{0ex}{0ex}}⇒10+\left(n-1\right)10=500\phantom{\rule{0ex}{0ex}}⇒\left(n-1\right)10=490\phantom{\rule{0ex}{0ex}}⇒n=50\phantom{\rule{0ex}{0ex}}Now,\phantom{\rule{0ex}{0ex}}{S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}{S}_{50}=\frac{50}{2}\left[20+\left(49\right)10\right]\phantom{\rule{0ex}{0ex}}=25\left[510\right]\phantom{\rule{0ex}{0ex}}=12750$
Hence, the sum of Integers from 1 and 500 which are multiples of 2 or 5 is
[ sum of all the multiples of 2 + sum of all the multiples of 5 ] − [ sum of all the multiples of 2 and 5 ]
= [ 62750 + 25250 ] − 12750
= 75250

#### Question 56:

Let there be an A.P. with first term 'a', common difference 'd'. If an denotes in nth term and Sn the sum of first n terms, find.

(i) n and Sn, if a = 5, d = 3 and an = 50.

(ii) n and a, if an = 4, d = 2 and Sn = −14.

(iii) d, if a = 3, n = 8 and Sn = 192.

(iv) a, if an = 28, Sn = 144 and n= 9.

(v) n and d, if a = 8, an = 62 and Sn = 210

(vi) n and an, if a= 2, d = 8 and Sn = 90.

(vii) .

(viii)

(i) Here, we have an A.P. whose nth term (an), first term (a) and common difference (d) are given. We need to find the number of terms (n) and the sum of first n terms (Sn).

Here,

First term (a) = 5

Last term () = 50

Common difference (d) = 3

So here we will find the value of n using the formula,

So, substituting the values in the above mentioned formula

Further simplifying for n,

Now, here we can find the sum of the n terms of the given A.P., using the formula,

Where, a = the first term

l = the last term

So, for the given A.P, on substituting the values in the formula for the sum of n terms of an A.P., we get,

Therefore, for the given A.P

(ii) Here, we have an A.P. whose nth term (an), sum of first n terms (Sn) and common difference (d) are given. We need to find the number of terms (n) and the first term (a).

Here,

Last term () = 4

Common difference (d) = 2

Sum of n terms (Sn) = −14

So here we will find the value of n using the formula,

So, substituting the values in the above mentioned formula

Now, here the sum of the n terms is given by the formula,

Where, a = the first term

l = the last term

So, for the given A.P, on substituting the values in the formula for the sum of n terms of an A.P., we get,

Equating (1) and (2), we get,

So, we get the following quadratic equation,

Further, solving it for a by splitting the middle term,

So, we get,

Or

Substituting, in (1),

Here, we get n as negative, which is not possible. So, we take,

Therefore, for the given A.P

(iii) Here, we have an A.P. whose first term (a), sum of first n terms (Sn) and the number of terms (n) are given. We need to find common difference (d).

Here,

First term () = 3

Sum of n terms (Sn) = 192

Number of terms (n) = 8

So here we will find the value of n using the formula,

So, to find the common difference of this A.P., we use the following formula for the sum

of n terms of an A.P

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 8, we get,

Further solving for d,

Therefore, the common difference of the given A.P. is.

(iv) Here, we have an A.P. whose nth term (an), sum of first n terms (Sn) and the number of terms (n) are given. We need to find first term (a).

Here,

Last term () = 28

Sum of n terms (Sn) = 144

Number of terms (n) = 9

Now,

Also, using the following formula for the sum of n terms of an A.P

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 9, we get,

Multiplying (1) by 9, we get

Further, subtracting (3) from (2), we get

Therefore, the first term of the given A.P. is.

(v) Here, we have an A.P. whose nth term (an), sum of first n terms (Sn) and first term (a) are given. We need to find the number of terms (n) and the common difference (d).

Here,

First term () = 8

Last term () = 62

Sum of n terms (Sn) = 210

Now, here the sum of the n terms is given by the formula,

Where, a = the first term

l = the last term

So, for the given A.P, on substituting the values in the formula for the sum of n terms of an A.P., we get,

Also, here we will find the value of d using the formula,

So, substituting the values in the above mentioned formula

Therefore, for the given A.P

(vi) Here, we have an A.P. whose first term (a), common difference (d) and sum of first n terms are given. We need to find the number of terms (n) and the nth term (an).

Here,

First term (a) = 2

Sum of first nth terms () = 90

Common difference (d) = 8

So, to find the number of terms (n) of this A.P., we use the following formula for the sum

of n terms of an A.P

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 8, we get,

Further solving the above quadratic equation,

Further solving for n,

Now,

Also,

Since n cannot be a fraction

Thus, n = 5

Also, we will find the value of the nth term (an) using the formula,

So, substituting the values in the above mentioned formula

Therefore, for the given A.P.

(vii)
${a}_{k}={S}_{k}-{S}_{k-1}\phantom{\rule{0ex}{0ex}}⇒164=\left(3{k}^{2}+5k\right)-\left(3{\left(k-1\right)}^{2}+5\left(k-1\right)\right)\phantom{\rule{0ex}{0ex}}⇒164=3{k}^{2}+5k-3{k}^{2}+6k-3-5k+5\phantom{\rule{0ex}{0ex}}⇒164=6k+2\phantom{\rule{0ex}{0ex}}⇒6k=162\phantom{\rule{0ex}{0ex}}⇒k=27\phantom{\rule{0ex}{0ex}}$

(viii) Given d = 22, ${a}_{22}=149$ ,n=22
We know that

$149=a+\left(22-1\right)22\phantom{\rule{0ex}{0ex}}149=a+462\phantom{\rule{0ex}{0ex}}a=-313\phantom{\rule{0ex}{0ex}}$

Now, Sum is given by

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 22, we get

${S}_{22}=\frac{22}{2}\left\{2×\left(-313\right)+\left(22-1\right)×22\right)\right\}\phantom{\rule{0ex}{0ex}}{S}_{22}=11\left\{-626+462\right\}\phantom{\rule{0ex}{0ex}}{S}_{22}=-1804$

Hence, the sum of 22 terms is −1804.

#### Question 57:

If Sn denotes the sum of first n terms of an A.P., prove that S12 = 3(S8 − S4).

Let a be the first term and d be the common difference.

We know that, sum of first n terms = Sn $\frac{n}{2}$[2a + (n − 1)d]

Now,
S$\frac{4}{2}$[2a + (4 − 1)d]
= 2(2a + 3d)
= 4a + 6d              ....(1)

S$\frac{8}{2}$[2a + (8 − 1)d]
= 4(2a + 7d)
= 8a + 28d            ....(2)

S12 $\frac{12}{2}$[2a + (12 − 1)d]
= 6(2a + 11d)
= 12a + 66d          ....(3)

On subtracting (1) from (2), we get
S8 − S8a + 28d − (4a + 6d)
4a + 22d

Multiplying both sides by 3, we get

3(S8 − S4) = 3(4a + 22d)
= 12a + 66d
= S
12                 [From (3)]

Thus, S12 = 3(S8 − S4).

#### Question 58:

A thief, after commiting a theft runs at a uniform speed of 50 m/minute . After 2 minutes, a policeman runs to catch him. He goes 60 m in first minute and increases his speed by 5 m/minute everey suceeding minute. After how many minutes, the policeman will catch the thief.?

Suppose the policeman catches the thief after t minutes.

Uniform speed of the thief = 50 m/min

∴ Distance covered by thief in (t + 2) minutes = 50 m/min × (t + 2) min = 50 (t + 2) m

The distance covered by the policeman in t minutes is in AP, with 60 and 5 as the first term and the common difference, respectively.

Now,

Distance covered by policeman in t minutes = Sum of t terms

When the policeman catches the thief, we have

$\frac{t}{2}\left[115+5t\right]=50\left(t+2\right)\phantom{\rule{0ex}{0ex}}115t+5{t}^{2}=100t+200\phantom{\rule{0ex}{0ex}}⇒5{t}^{2}+15t-200=0\phantom{\rule{0ex}{0ex}}⇒{t}^{2}+3t-40=0\phantom{\rule{0ex}{0ex}}⇒\left(t+8\right)\left(t-5\right)=0$

So, t = −8 or t = 5

∴ t = 5  (As t cannot be negative)

Thus, the policeman catches the thief after 5 min.

#### Question 59:

The sums of first n terms of three A.P. are S1,S2 and S3. The first term of each is 5 and their common differences are 2, 4 and 6 respectively. Prove that S1 + S3 = 2S2

Hence proved.

#### Question 60:

Resham wanted to save at least ₹6500 for sending her daughter to school next year (after 12 months). She saved ₹450 in the first month and raised her savings by ​₹20 every next month. How much will she able to save in next 12 months ? Will she be able to send her daughter to the school next year ?

It is given that Reshma saved ₹450 in the first month and raised her savings by ₹20 every next month.
So, her savings are in an AP, with the first term (a) = ₹450 and the common difference (d) = ₹20.
We need to find her savings for 12 months, so n = 12.
We know that the sum of n terms of an AP is ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$.
Reshma's savings for 12 months:

So, she will save ₹6,720 in 12 months.
She needed to save at least ₹6,500 for sending her daughter to school next year.
Since ₹6,720 is greater than ₹6,500, Reshma can send her daughter to school.
The question aims to encourage personal savings and emphasise the need of female education.

#### Question 61:

In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two sections, find how many trees were planted by the students.

Let a be the first term and d be the common difference.

We know that, sum of first n terms = Sn $\frac{n}{2}$[2a + (n − 1)d]

It can be observed that the number of trees planted by the students forms an A.P.
2, 4, 6, 8, ... , 24

Here, a = 2, d = 2 and n = 12.

$\therefore {S}_{n}=\frac{12}{2}\left[2×2+\left(12-1\right)2\right]$
= 6(4 + 22)
= 6(26)
= 156

Therefore, trees planted by 1 section of all the classes = 156.
Number of trees planted by 2 sections of all the classes = 2 × 156 = 312

Thus, 312 trees ​were planted by the students.

#### Question 62:

Ramkali would need ₹1800 for admission fee and books etc., for her daughter to start going to school from next year. She saved ₹50 in the first month of this year and increased her monthly saving by ₹20. After a year, how much money will she save? Will she be able to fulfil her dream of sending her daughter to school?

Let a be the first term and d be the common difference.

We know that, sum of first n terms = Sn $\frac{n}{2}$[2a + (n − 1)d]

According to the question,
Saving of Ramkali in 1 year = ₹50 + ₹70 + ₹90.......

Here, a = 50, d = 70 − 50 = 20 and n = 12.

∴ S12 = $\frac{12}{2}$[2 × 50 + (12 − 1)20]
= 6[100 + 220]
= 6 × 320
= 1920

Hence, After a year, she will save ₹1920.

Since, required amount for admission is ₹1800 and her savings will be ₹1920.

Thus, yes she will be able to fulfil her dream of sending her daughter to school.

#### Question 63:

A man saved Rs 16500 in ten years. In each year after the first he saved Rs 100 more than he did in the preceding year. How much did he save in the first year?

Here, we are given that the total saving of a man is Rs 16500 and every year he saved Rs 100 more than the previous year.

So, let us take the first installment as a.

Second installment =

Third installment =

So, these installments will form an A.P. with the common difference (d) = 100

The sum of his savings every year

Number of years (n) = 10

So, to find the first installment, we use the following formula for the sum of n terms of an A.P.,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 10, we get,

Further solving for a,

Therefore, man saved in the first year.

#### Question 64:

A man saved Rs. 32 during the first year, Rs 36 in the second year and in this way he increases his saving by Rs 4 every year. Find in what time his saving will be Rs. 200.

Here, we are given that the total saving of a man is Rs 200. In the first year he saved Rs 32 and every year he saved Rs 4 more than the previous year.

So, the first installment = 32.

Second installment = 36

Third installment =

So, these installments will form an A.P. with the common difference (d) = 4

The sum of his savings every year

We need to find the number of years. Let us take the number of years as n.

So, to find the number of years, we use the following formula for the sum of n terms of an A.P.,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 10, we get,

Further solving for n by splitting the middle term, we get,

So,

Or

Since number of years cannot be negative. So, in, his savings will be Rs 200.

#### Question 65:

A man arranges to pay off a dept of Rs 3600 by 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of all debt unpaid, find the value of the first instalment.

In the given problem,

Total amount of debt to be paid in 40 installments

After 30 installments one−third of his debt is left unpaid. This means that he paid two third of the debt in 30 installments. So,

Amount he paid in 30 installments

Let us take the first installment as a and common difference as d.

So, using the formula for the sum of n terms of an A.P,

Let us find a and d, for 30 installments.

Similarly, we find a and d for 40 installments.

Subtracting (1) from (2), we get,

Further solving for d,

Substituting the value of d in (1), we get.

Therefore, the first installment is.

#### Question 66:

There are 25 trees at equal distances of 5 metres in a line with a well, the distance of the well from the nearest tree being 10 metres. A gardener water all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to eater all the trees.

In the given problem, there are 25 trees in a line with a well such that the distance between two trees is 5 meters and the distance between the well and the first tree is 10 meters.

So, the total distance covered to water first tree

Then he goes back to the well to get water.

So,

The total distance covered to water second tree

The total distance covered to water third tree

The total distance covered to water fourth tree

So, from second tree onwards, the distance covered by the gardener forms an A.P. with the first term as 25 and common difference as 10.

So, the total distance covered for 24 trees can be calculated by using the formula for the sum of n terms of an A.P,

We get,

So, while watering the 24 trees he covered 3360 meters. Also, to water the first tree he covers 10 meters. So the distance covered while watering 25 trees is 3370 meters.

Now, the distance between the last tree and the well

So, to get back to the well he covers an additional 130 m. Therefore, the total distance covered by the gardener

Therefore, the total distance covered by the gardener is.

#### Question 67:

A man is employed to count Rs 10710. He counts at the rate of Rs 180 per minute for half an hour. After this he counts at the rate of Rs 3 less every minute than the preceding minute. Find the time taken by him to count the entire amount.

In the given problem, the total amount = Rs 10710.

For the first half and hour (30 minutes) he counts at a rate of Rs 180 per minute. So,

The amount counted in 30 minutes

So, amount left after half an hour

After 30 minutes he counts at a rate of Rs 3 less every minute. So,

At 31st minute the rate of counting per minute = 177.

At 32nd minute the rate of counting per minute = 174.

So, the rate of counting per minute for each minute will form an A.P. with the first term as 177 and common difference as −3.

So, the total time taken to count the amount left after half an hour can be calculated by using the formula for the sum of n terms of an A.P,

We get,

So, we get the following quadratic equation,

Solving the equation by splitting the middle term, we get,

So,

Or

Therefore, the total time required for counting the entire amount

So, the total time required for counting the entire amount is.

#### Question 68:

A piece of equipment cost a certain factory Rs 60,000. If it depreciates in value, 15% the first, 13.5% the next year, 12% the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost?

In the given problem,

Cost of the equipment = Rs 600,000

It depreciates by 15% in the first year. So,

Depreciation in 1 year

It depreciates by 13.5% of the original cost in the 2 year. So,

Depreciation in 2 year

Further, it depreciates by 12% of the original cost in the 3 year. So,

Depreciation in 3 year

So, the depreciation in value of the equipment forms an A.P. with first term as 90000 and common difference as −9000.

So, the total depreciation in value in 10 years can be calculated by using the formula for the sum of n terms of an A.P,

We get,

So, the total depreciation in the value after 10 years is Rs 495000.

Therefore, the value of equipment

So, the value of the equipment after 10 years is.

#### Question 69:

A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each prize.

In the given problem,

Total amount of money (Sn) = Rs 700

There are a total of 7 prizes and each prize is Rs 20 less than the previous prize. So let us take the first prize as Rs a.

So, the second prize will be Rs, third prize will be Rs.

Therefore, the prize money will form an A.P. with first term a and common difference −20.

So, using the formula for the sum of n terms,

We get,

On further simplification, we get,

Therefore, the value of first prize is Rs 160.

Second prize = Rs 140

Third prize = Rs 120

Fourth prize = Rs 100

Fifth prize = Rs 80

Sixth prize = Rs 60

Seventh prize= Rs 40

So the values of prizes are

#### Question 70:

If Sn denotes the sum of the first n terms of an A.P., prove that S30 = 3(S20 − S10)

Let a be the first term and d be the common difference.

We know that, sum of first n terms = Sn $\frac{n}{2}$[2a + (n − 1)d]

Now,
S10 $\frac{10}{2}$[2a + (10 − 1)d]
= 5(2a + 9d)
= 10a + 45d          ....(1)

S20 $\frac{20}{2}$[2a + (20 − 1)d]
= 10(2a + 19d)
= 20a + 190d        ....(2)

S30 $\frac{30}{2}$[2a + (30 − 1)d]
= 15(2a + 29d)
= 30a + 435d        ....(3)

On subtracting (1) from (2), we get
S20 − S10 = 20a + 190d − (10a + 45d)
= 10a + 145d

On multiplying both sides by 3, we get
3(S20 − S10) = 3(10a + 145d)
= 30a + 435d
= S
30                   [From (3)]

Hence, S30 = 3(S20 − S10)

#### Question 71:

Solve the question

Suppose x is nth term of the given A.P.

#### Question 72:

Which term of the A.P.  will be $-77$ ? Find the sum of this A.P. upto the term $-77$ .

Here, a = −2 and d = −5.

#### Question 73:

The sum of first n terms of an A.P. whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another A.P. whose first term is $-30$ and common difference is 8 . Find n.

According to the question, we have
$\frac{n}{2}\left[2\left(8\right)+\left(n-1\right)20\right]=\frac{2n}{2}\left[2\left(-30\right)+\left(2n-1\right)8\right]\phantom{\rule{0ex}{0ex}}⇒\left[16+20n-20\right]=2\left[-60+16n-8\right]\phantom{\rule{0ex}{0ex}}⇒20n-4=-136+32n\phantom{\rule{0ex}{0ex}}⇒12n=132\phantom{\rule{0ex}{0ex}}⇒n=11$

#### Question 74:

The students of a school decided to beautify the school on the annual day by fixing colourful on the straight passage of the school . They have 27 flags to be fixed at intervals of every 2 metre . The flags are stored at the position of the middle most flag . Ruchi was given the responsibility of placing the flags . Ruchi kept her books where the flags were stored . She could carry only one flag at a time . How much distance did she cover in completing this job and returning back to collect her books ? What is the maximum distance she travelled carrying a flag ?

Distance covered to place the first flag to the left of the middle flag = 2 × 2 m = 4 m.
Distance covered to place the second flag to the left of the middle flag = 2 × 4 m = 8 m.
Similarly,
Distance covered to place the thirteenth flag to the left of the middle flag = 2 × 26 m = 52 m.
Now,
The total distance covered = 2 ( 4 + 8 + 12 +....+ 52)
The sum is as follows:

The total distance trvelled is 728 m and maximum distance travelled in carrying a flag is 26 m.

#### Question 1:

Mark the correct alternative in each of the following:

If 7th and 13th terms of an A.P. be 34 and 64 respectively, then its 18th term is

(a) 87

(b) 88

(c) 89

(d) 90

In the given problem, we are given 7th and 13th term of an A.P.

We need to find the 26th term

Here,

Now, we will find and using the formula

So,

Also,

Further, to solve for a and d

On subtracting (1) from (2), we get

Substituting (3) in (1), we get

Thus,

So, for 18th term (n = 18),

Substituting the above values in the formula,

Therefore,

Hence, the correct option is (c).

#### Question 2:

If the sum of P terms of an A.P. is q and the sum of q terms is p, then the sum of p + q terms will be

(a) 0

(b) pq

(c) p + q

(d) −(p + q)

In the given problem, we are given and

We need to find

Now, as we know,

So,

Similarly,

Subtracting (2) from (1), we get

Now,

Thus,

Hence, the correct option is (d).

#### Question 3:

If the sum of n terms of an A.P. be 3n2 + n and its common difference is 6, then its first term is

(a) 2

(b) 3

(c) 1

(d) 4

In the given problem, the sum of n terms of an A.P. is given by the expression,

Here, we can find the first term by substitutingas sum of first term of the A.P. will be the same as the first term. So we get,

Therefore, the first term of this A.P is. So, the correct option is (d).

#### Question 4:

The first and last terms of an A.P. are 1 and 11. If the sum of its terms is 36, then the number of terms will be

(a) 5

(b) 6

(c) 7

(d) 8

In the given problem, we need to find the number of terms in an A.P. We are given,

First term (a) = 1

Last term (an) = 11

Sum of its terms

Now, as we know,

Where, a = the first term

l = the last term

So, we get,

Therefore, the total number of terms in the given A.P. is

Hence the correct option is (b).

#### Question 5:

If the sum of n terms of an A.P. is 3n2 + 5n then which of its terms is 164?

(a) 26th

(b) 27th

(c) 28th

(d) none of these.

Here, the sum of first n terms is given by the expression,

We need to find which term of the A.P. is 164.

Let us take 164 as the nth term

So we know that the nthterm of an A.P. is given by,

So,

Using the property,

We get,

Further solving for n, we get

Therefore,

Hence the correct option is (b).

#### Question 6:

If the sum of n terms of an A.P. is 2n2 + 5n, then its nth term is

(a) 4n − 3

(b) 3n − 4

(c) 4n + 3

(d) 3n + 4

Here, the sum of first n terms is given by the expression,

We need to find the nth term.

So we know that the nthterm of an A.P. is given by,

So,

Using the property,

We get,

Therefore,

Hence the correct option is (c).

#### Question 7:

If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third of these terms is 273, then the third term is

(a) 13

(b) 9

(c) 21

(d) 17

In the given problem, the sum of three consecutive terms of an A.P is 51 and the product of the first and the third terms is 273.

We need to find the third term.

Here,

Let the three terms be where, a is the first term and d is the common difference of the A.P

So,

Also,

Further solving for d,

Now, it is given that this is an increasing A.P. so d cannot be negative.

So, d = 4

Substituting the values of a and d in the expression for the third term, we get,

Third term =

So,

Therefore, the third term is

Hence, the correct option is (c).

#### Question 8:

If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times, the least, then the numbers are

(a) 5, 10, 15, 20

(b) 4, 101, 16, 22

(c) 3, 7, 11, 15

(d) none of these

Here, we are given that four numbers are in A.P., such that their sum is 50 and the greatest number is 4 times the smallest.

So, let us take the four terms as.

Now, we are given that sum of these numbers is 50, so we get,

Also, the greatest number is 4 times the smallest, so we get,

Now, using (2) in (1), we get,

Now, using the value of a in (2), we get,

So, first term is given by,

Second term is given by,

Third term is given by,

Fourth term is given by,

Therefore, the four terms are

Hence, the correct option is (a).

#### Question 9:

Let Sn denote the sum of n terms of an A.P. whose first term is a. If the common difference d is given by = Sn − kSn−1 + Sn−2, then k =

(a) 1

(b) 2

(c) 3

(d) none of these.

In the given problem, we are given

We need to find the value of k

So here,

First term = a

Common difference = d

Sum of n terms = Sn

Now, as we know,

Also, for n-1 terms,

Further, for n-2 terms,

Now, we are given,

Using (1), (2) and (3) in the given equation, we get

Taking common, we get,

Taking 2 common from the numerator, we get,

Therefore,

Hence, the correct option is (b).

#### Question 10:

The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is given by $\frac{{l}^{2}-{a}^{2}}{k-\left(l+a\right)}$, then k =

In the given problem, we are given the first, last term, sum and the common difference of an A.P.

We need to find the value of k

Here,

First term = a

Last term = l

Sum of all the terms = S

Common difference (d) =

Now, as we know,

Further, substituting (1) in the given equation, we get

Now, taking d in common, we get,

Taking (n-1) as common, we get,

Further, multiplying and dividing the right hand side by 2, we get,

Now, as we know,

Thus,

Therefore, the correct option is (b).

#### Question 11:

If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers, then k =

(a) $\frac{1}{n}$

(b) $\frac{n-1}{n}$

(c) $\frac{n+1}{2n}$

(d) $\frac{n+1}{n}$

In the given problem, we are given that the sum of the first n even natural numbers is equal to k times the sum of first n odd natural numbers.

We need to find the value of k

Now, we know that the first odd natural number is 1. Also, all the odd terms will form an A.P. with the common difference of 2.

So here,

First term (a) = 1

Common difference (d) = 2

So, let us take the number of terms as n

Now, as we know,

So, for n terms,

Also, we know that the first even natural number is 2. Also, all the odd terms will form an A.P. with the common difference of 2.

So here,

First term (a) = 2

Common difference (d) = 2

So, let us take the number of terms as n

So, for n terms,

Solving further, we get

Now, as the sum of the first n even natural numbers is equal to k times the sum of first n odd natural numbers

Using (1) and (2), we get

Therefore,

Hence, the correct option is (d).

#### Question 12:

If the first, second and last term of an A.P. are a, b and 2a respectively, its sum is

(a) $\frac{ab}{2\left(b-a\right)}$

(b) $\frac{ab}{\left(b-a\right)}$

(c) $\frac{3ab}{2\left(b-a\right)}$

(d) none of these

In the given problem, we are given first, second and last term of an A.P. We need to find its sum.

So, here

First term = a

Second term (a2) = b

Last term (l) = 2a

Now, using the formula

Also,

Further as we know,

Substituting (2) in the above equation, we get

Using (1), we get

Thus,

Therefore, the correct option is (c).

#### Question 13:

If S1 is the sum of an arithmetic progression of 'n' odd number of terms and S2 the sum of the terms of the series in odd places, then $\frac{{S}_{1}}{{S}_{2}}=$

(a) $\frac{2n}{n+1}$

(b) $\frac{n}{n+1}$

(c) $\frac{n+1}{2n}$

(d) $\frac{n+1}{n}$

In the given problem, we are given as the sum of an A.P of ‘n’ odd number of terms and the sum of the terms of the series in odd places.

We need to find

Now, let a1, a2…. an be the n terms of A.P

Where n is odd

Let d be the common difference of the A.P

Then,

And be the sum of the terms of the places in odd places,

Where, number of terms =

Common difference = 2d

So,

Now,

Thus,

Therefore, the correct option is (a).

#### Question 14:

If in an A.P. Sn = n2p and Sm = m2p, where Sr denotes the sum of r terms of the A.P., then Sp is equal to

(a) $\frac{1}{2}{p}^{3}$

(b) m n p

(c) p3

(d) (m + n) p2

In the given problem, we are given an A.P whose and

We need to find

Now, as we know,

Where, first term = a

Common difference = d

Number of terms = n

So,

Similarly,

Equating (1) and (2), we get,
$\frac{1}{2n}\left(2a+nd-d\right)=\frac{1}{2m}\left(2a+md-d\right)$
$⇒m\left(2a+nd-d\right)=n\left(2a+md-d\right)$
$⇒2am+mnd-md=2an+mnd-nd$

Solving further, we get,

Further, substituting (3) in (1), we get,

Now,

Thus,

Therefore, the correct option is (C).

#### Question 15:

If Sn denote the sum of the first n terms of an A.P. If S2n = 3Sn, then S3n : Sn is equal to

(a) 4

(b) 6

(c) 8

(d) 10

Here, we are given an A.P. whose sum of n terms is Sn and.

We need to find.

Here we use the following formula for the sum of n terms of an A.P.,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, first we find S3n,

Similarly,

Also,

Now,

So, using (2) and (3), we get,

On further solving, we get,

So,

Taking common, we get,

Therefore,

Hence, the correct option is (b).

#### Question 16:

In an AP. Sp = q, Sq = p and Sr denotes the sum of first r terms. Then, Sp+q is equal to

(a) 0

(b) −(p + q)

(c) p + q

(d) pq

In the given problem, we are given and

We need to find

Now, as we know,

So,

Similarly,

Subtracting (2) from (1), we get

Now,

Thus,

Hence, the correct option is (b).

#### Question 17:

If Sr denotes the sum of the first r terms of an A.P. Then , S3n: (S2nSn) is

(a) n

(b) 3n

(c) 3

(d) none of these

Here, we are given an A.P. whose sum of r terms is Sr. We need to find.

Here we use the following formula for the sum of n terms of an A.P.,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, first we find S3n,

Similarly,

Also,

So, using (1), (2) and (3), we get,

Taking common, we get,

Therefore,

Hence, the correct option is (c).

#### Question 18:

If the first term of an A.P. is 2 and common difference is 4, then the sum of its 40 terms is
(a) 3200
(b) 1600
(c) 200
(d) 2800

In the given problem, we need to find the sum of 40 terms of an arithmetic progression, where we are given the first term and the common difference. So, here we use the following formula for the sum of n terms of an A.P.,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

Given,

First term (a) = 2

Common difference (d) = 4

Number of terms (n) = 40

So, using the formula we get,

Therefore, the sum of first 40 terms for the given A.P. is. So, the correct option is (a).

#### Question 19:

The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is

(a) 5

(b) 10

(c) 12

(d) 14

(e) 20

In the given problem, we have an A.P.

Here, we need to find the number of terms n such that the sum of n terms is 406.

So here, we will use the formula,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

The first term (a) = 3

The sum of n terms (Sn) = 406

Common difference of the A.P. (d) =

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

So, we get the following quadratic equation,

On solving by splitting the middle term, we get,

Further,

Or,

Since, the number of terms cannot be a fraction, the number of terms (n) is

Hence, the correct option is (d).

#### Question 20:

Sum of n terms of the series $\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+..is$

(a)

(b)

(c)

(d) 1

In the given problem, we need to find the sum of terms for a given arithmetic progression,

So, here we use the following formula for the sum of n terms of an A.P.,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

Here,

Common difference of the A.P. (d) =

Number of terms (n) = n

First term for the given A.P. (a) =

So, using the formula we get,

Now, taking common from both the terms inside the bracket we get,

Therefore, the sum of first n terms for the given A.P. is. So, the correct option is (c).

#### Question 21:

The 9th term of an A.P. is 449 and 449th term is 9. The term which is equal to zero is

(a) 501th

(b) 502th

(c) 508th

(d) none of these

In the given problem, let us take the first term as a and the common difference as d.

Here, we are given that,

We need to find n

Also, we know,

For the 9th term (n = 9),

Similarly, for the 449th term (n = 449),

Subtracting (3) from (4), we get,

Now, to find a, we substitute the value of d in (3),

So, for the given A.P

So, let us take the term equal to zero as the nth term. So,

So,

Therefore, the correct option is (d).

#### Question 22:

If $\frac{1}{x+2},\frac{1}{x+3},\frac{1}{x+5}$ are in A.P. Then, x =

(a) 5

(b) 3

(c) 1

(d) 2

Here, we are given three terms,

First term (a1) =

Second term (a2) =

Third term (a3) =

We need to find the value of x for which these terms are in A.P. So, in an A.P. the difference of two adjacent terms is always constant. So, we get,

Also,

Now, on equating (1) and (2), we get,

Therefore, for, these three terms will form an A.P.

So, the correct option is (c).

#### Question 23:

The nth term of an A.P., the sum of whose n terms is Sn, is

(a) Sn + Sn−1

(b) Sn − Sn−1

(c) Sn + Sn+1

(d) Sn − Sn+1

A.P. we use following formula,

So, the nth term of the A.P. is given by . Therefore, the correct option is (b).

#### Question 24:

The common difference of an A.P., the sum of whose n terms is Sn, is

(a) Sn − 2Sn−1 + Sn−2

(b) Sn − 2Sn−1 − Sn−2

(c) Sn − Sn−2

(d) Sn − Sn−1

Here, we are given an A.P. the sum of whose n terms is Sn. So, to calculate the common difference of the A.P, we find two consecutive terms of the A.P.

Now, the nth term of the A.P will be given by the following formula,

Next, we find the (n − 1)th term using the same formula,

Now, the common difference of an A.P. (d) =

Therefore,

Hence the correct option is (a).

#### Question 25:

If the sums of n terms of two arithmetic progressions are in the ratio $\frac{3n+5}{5n-7}$, then their nth terms are in the ratio

(a) $\frac{3n-1}{5n-1}$

(b) $\frac{3n+1}{5n+1}$

(c) $\frac{5n+1}{3n+1}$

(d) $\frac{5n-1}{3n-1}$

In the given problem, the ratio of the sum of n terms of two A.P’s is given by the expression,

We need to find the ratio of their nth terms.

Here we use the following formula for the sum of n terms of an A.P.,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So,

Where, a and d are the first term and the common difference of the first A.P.

Similarly,

Where, a and d are the first term and the common difference of the first A.P.

So,

Equating (1) and (2), we get,

Now, to find the ratio of the nth term, we replace n by. We get,

As we know,

Therefore, we get,

Hence the correct option is (b).

#### Question 26:

If Sn denote the sum of n terms of an A.P. with first term a and common difference d such that $\frac{Sx}{Skx}$ is independent of x, then

(a) d= a

(b) d = 2a

(c) a = 2d

(d) d = −a

Here, we are given an A.P. with a as the first term and d as the common difference. The sum of n terms of the A.P. is given by Sn.

We need to find the relation between a and d such thatis independent of

So, let us first find the values of Sx and Skx using the following formula for the sum of n terms of an A.P.,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, we get,

Similarly,

So,

Now, to get a term independent of x we have to eliminate the other terms, so we get

So, if we substitute, we get,

Therefore,

Hence, the correct option is (b).

#### Question 27:

If the first term of an A.P. is a and nth term is b, then its common difference is

(a) $\frac{b-a}{n+1}$

(b) $\frac{b-a}{n-1}$

(c) $\frac{b-a}{n}$

(d) $\frac{b+a}{n-1}$

Here, we are given the first term of the A.P. as a and the nth term (an) as b. So, let us take the common difference of the A.P. as d.

Now, as we know,

On substituting the values given in the question, we get.

Therefore,

Hence the correct option is (b).

#### Question 28:

The sum of first n odd natural numbers is

(a) 2n − 1

(b) 2n + 1

(c) n2

(d) n2 − 1

In this problem, we need to find the sum of first n odd natural numbers.

So, we know that the first odd natural number is 1. Also, all the odd terms will form an A.P. with the common difference of 2.

So here,

First term (a) = 1

Common difference (d) = 2

So, let us take the number of terms as n

Now, as we know,

So, for n terms,

Therefore, the sum of first n odd natural numbers is.

Hence the correct option is (c).

#### Question 29:

Two A.P.'s have the same common difference. The first term of one of these is 8 and that of the other is 3. The difference between their 30th term is

(a) 11

(b) 3

(c) 8

(d) 5

Here, we are given two A.P.’s with same common difference. Let us take the common difference as d.

Given,

First term of first A.P. (a) = 8

First term of second A.P. (a) = 3

We need to find the difference between their 30th terms.

So, let us first find the 30th term of first A.P.

Similarly, we find the 30th term of second A.P.

Now, the difference between the 30th terms is,

Therefore,

Hence, the correct option is (d).

#### Question 30:

If 18, a, b, −3 are in A.P., the a + b =

(a) 19

(b) 7

(c) 11

(d) 15

Here, we are given four terms which are in A.P.,

First term (a1) =

Second term (a2) =

Third term (a3) =

Fourth term (a4)=

So, in an A.P. the difference of two adjacent terms is always constant. So, we get,

Also,

Now, on equating (1) and (2), we get,

Therefore,

Hence the correct option is (d).

#### Question 31:

The sum of n terms of two A.P.'s are in the ratio 5n + 9 : 9n + 6. Then, the ratio of their 18th term is

(a) $\frac{179}{321}$

(b) $\frac{178}{321}$

(c) $\frac{175}{321}$

(d) $\frac{176}{321}$

In the given problem, the ratio of the sum of n terms of two A.P’s is given by the expression,

We need to find the ratio of their 18th terms.

Here we use the following formula for the sum of n terms of an A.P.,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So,

Where, a and d are the first term and the common difference of the first A.P.

Similarly,

Where, a’ and d are the first term and the common difference of the first A.P.

So,

Equating (1) and (2), we get,

Now, to find the ratio of the nth term, we replace n by. We get,

As we know,

Therefore, for the 18th terms, we get,

Hence

Hence no option is correct.

#### Question 32:

If then n =

(a) 8

(b) 7

(c) 10

(d) 11

Here, we are given,

We need to find n.

So, first let us find out the sum of n terms of the A.P. given in the numerator. Here we use the following formula for the sum of n terms of an A.P.,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

Here,

Common difference of the A.P. (d) =

Number of terms (n) = n

First term for the given A.P. (a) =

So, using the formula we get,

Similarly, we find out the sum of terms of the A.P. given in the denominator.

Here,

Common difference of the A.P. (d) =

Number of terms (n) = n

First term for the given A.P. (a) =

So, using the formula we get,

Now substituting the values of (2) and (3) in equation (1), we get,

Further solving the quadratic equation for n by splitting the middle term, we get,

So, we get

Or

Since n is a whole number, it cannot be a fraction. So,

Therefore, the correct option is (b).

#### Question 33:

The sum of n terms of an A.P. is 3n2 + 5n, then 164 is its

(a) 24th term

(b) 27th term

(c) 26th term

(d) 25th term

Here, the sum of first n terms is given by the expression,

We need to find which term of the A.P. is 164.

Let us take 164 as the nth term.

So we know that the nthterm of an A.P. is given by,

So,

Using the property,

We get,

Further solving for n, we get

Therefore,

Hence the correct option is (b).

#### Question 34:

If the nth term of an A.P. is 2n + 1, then the sum of first n terms of the A.P. is

(a) n(n − 2)

(b) n(n + 2)

(c) n(n + 1)

(d) n(n − 1)

Here, we are given an A.P. whose nth term is given by the following expression, . We need to find the sum of first n terms.

So, here we can find the sum of the n terms of the given A.P., using the formula,

Where, a = the first term

l = the last term

So, for the given A.P,

The first term (a) will be calculated using in the given equation for nth term of A.P.

Now, the last term (l) or the nth term is given

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

Therefore, the sum of the n terms of the given A.P. is. So the correct option is (b).

#### Question 35:

If 18th and 11th term of an A.P. are in the ratio 3 : 2, then its 21st and 5th terms are in the ratio

(a) 3 : 2

(b) 3 : 1

(c) 1 : 3

(d) 2 : 3

In the given problem, we are given an A.P whose 18th and 11th term are in the ratio 3:2

We need to find the ratio of its 21st and 5th terms

Now, using the formula

,

Where,

a = first tem of the A.P

n = number of terms

d = common difference of the A.P

So,

Also,

Thus,

Further solving for a, we get

Now,

Also,

So,

Using (1) in the above equation, we get

Thus, the ratio of the 21st and 5th term is

Therefore the correct option is (b).

#### Question 36:

The sum of first 20 odd natural numbers is

(a) 100
(b) 210
(c) 400
(d) 420

Let a be the first term and d be the common difference.

We know that, sum of first n terms = Sn $\frac{n}{2}$[2a + (n − 1)d]

The given series is 1 + 3 + 5 + ......

First term = =  1.

Common difference = d = 3 − 1 = 2

∴ S20 $\frac{20}{2}$[2 × 1 + (20 − 1)2]
= 10(2 + 19 × 2)
= 10(40)
= 400

Hence, the correct option is (c).

#### Question 37:

The common difference of the A.P. is is

(a) −1
(b) 1
(c) q
(d) 2q

Let a be the first term and d be the common difference.

The given A.P. is

Common difference = d = Second term − First term
= $\frac{1-2q}{2q}-\frac{1}{2q}$
= $\frac{-2q}{2q}=-1$

Hence, the correct option is (a).

#### Question 38:

The common difference of the A.P. is

(a) $\frac{1}{3}$
(b) $-\frac{1}{3}$
(c) −b
(d) b

Let a be the first term and d be the common difference.

The given A.P. is

Common difference = d = Second term − First term
= $\frac{1-3b}{3}-\frac{1}{3}$
= $\frac{-3b}{3}=-b$

​Hence, the correct option is (c).

#### Question 39:

The common difference of the A.P. is

(a) 2b
(b) −2b
(c) 3
(d) −3

Let a be the first term and d be the common difference.

The given A.P. is

Common difference = d = Second term − First term
= $\frac{1-6b}{2b}-\frac{1}{2b}$
= $\frac{-6b}{2b}=-3$

Hence, the correct option is (d).

#### Question 40:

If k, 2k − 1 and 2k + 1 are three consecutive terms of an A.P., the value of k is

(a) −2
(b) 3
(c) −3
(d) 6

Since, k, 2k − 1 and 2k + 1 are three consecutive terms of an A.P.

Then, Second term − First term = Third term − Second term = d (common difference)
⇒ 2k − 1 − k = 2k + 1 − (2k − 1)
⇒ k − 1 = 2k + 1 − 2k + 1
⇒ k − 1 = 2
⇒ k = 2 + 1
⇒ k = 3

Hence, the correct option is (b).

#### Question 41:

The next term of the A.P.

(a) $\sqrt{70}$
(b) $\sqrt{84}$
(c) $\sqrt{97}$
(d) $\sqrt{112}$

Let a be the first term and d be the common difference.

The given A.P. is
i.e.,
i.e.,

Common difference = d = Second term − First term
= $2\sqrt{7}-\sqrt{7}$
= $\sqrt{7}$

∴ Next term of the A.P. = $3\sqrt{7}+\sqrt{7}$
= $4\sqrt{7}$
= $\sqrt{16×7}$
= $\sqrt{112}$

​Hence, the correct option is (d).

#### Question 42:

The first three terms of an A.P. respectively are 3y − 1, 3y + 5 and 5y + 1. Then, y equals

(a) −3
(b) 4
(c) 5
(d) 2

​Since, 3y − 1, 3y + 5 and 5y + 1 are first three terms of an A.P.

Then, Second term − First term = Third term − Second term = d (common difference)
⇒ 3y + 5 − (3y − 1) = 5y + 1 − (3y + 5)
⇒ 3y + 5 − 3y + 1 = 5y + 1 − 3y − 5
⇒ 6 = 2y − 4
⇒ 2y = 6 + 4
⇒ 2y = 10
y = 5

Hence, the correct option is (c).

#### Question 43:

The sum of first five multiples of 3 is
(a) 45
(b) 55
(c) 65
(d) 75

First five multiples of 3 are:
3, 6, 9, 12, 15

They are in A.P. with
First term = a = 3
Common difference = d = 6 − 3 = 3

​Hence, the correct option is (a).

#### Question 44:

The sum of first 16 terms of the A.P.: 10, 6, 2, ..., is
(a) –320
(b) 320
(c) –352
(d) –400

The given A.P. is:
10, 6, 2, ...,

First term = a = 10
Common difference = d = 6 − 10 = −4

​Hence, the correct option is (a).

#### Question 45:

If the first term of an A.P. is –5 and the common difference is 2, then the sum of first 6 terms is
(a) 0
(b) 5
(c) 6
(d) 15

Given:
First term = a = −5
Common difference = d = 2

​Hence, the correct option is (a).

#### Question 46:

The 4th term from the end of the AP: –11, –8, –5, ..., 49 is
(a) 37
(b) 40
(c) 43
(d) 58

The given A.P. is: –11, –8, –5, ..., 49

Rearranging the terms from last to first, we get
49, 46, 43, ..... –11

First term = a = 49
Common difference = d = 46 – 49
=  –3

The fourth term is:

Therefore, the 4th term from the end of the AP: –11, –8, –5, ..., 49 is 40.

​Hence, the correct option is (b).

#### Question 47:

Which term of the A.P. 21, 42, 63, 84, ... is 210?
(a) 9th
(b) 10th
(c) 11th
(d) 12th

The given A.P. is: 21, 42, 63, 84, ...

First term = a = 21
Common difference = d = 42 – 21
=  21
nth term = an = 210

Now,
${a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}⇒210=21+\left(n-1\right)\left(21\right)\phantom{\rule{0ex}{0ex}}⇒210=21+21n-21\phantom{\rule{0ex}{0ex}}⇒210=21n\phantom{\rule{0ex}{0ex}}⇒n=10$

Therefore, the 10th term of the A.P. 21, 42, 63, 84, ... is 210.

​Hence, the correct option is (b).

#### Question 48:

If the 2nd term of an A.P, is 13 and 5th term is 25, what is its 7th term?
(a) 30
(b) 33
(c) 37
(d) 38

Given:
2nd term = a2 = 13
5th term = a5 = 25

Now,

Therefore, the 7th term is 33.

​Hence, the correct option is (b).

#### Question 49:

The value of x for which 2x, x + 10 and 3x + 2 are the three consecutive terms of an A.P; is
(a) 6
(b) –6
(c) 18
(d) –18

Given  are the consecutive terms of an AP.
Therefore, the common difference will be same.
$⇒\left(x+10\right)-2x=\left(3x+2\right)-\left(x+10\right)\phantom{\rule{0ex}{0ex}}⇒x+10-2x=3x+2-x-10\phantom{\rule{0ex}{0ex}}⇒10-x=2x-8\phantom{\rule{0ex}{0ex}}⇒3x=18\phantom{\rule{0ex}{0ex}}⇒x=6$
Hence, the correct answer is option (a).

#### Question 50:

The first term of an A.P. is p and the common difference is q, then its 10th term is
(a) q + 9p
(b) p – 9q
(c) p + 9q
(d) 2p + 9q

The ${n}^{th}$ term of an AP where $a$ and $d$ are the first term and common difference respectively.
Therefore, ${10}^{th}$ term .

Hence, the correct answer is option (c).

#### Question 1:

The sequence $\frac{2a-6b}{3b},\frac{2a-3b}{3b},\frac{2a}{3b},\frac{2a+3b}{3b},\frac{2a+6b}{3b}$.... is an A.P. with common difference ________

The given A.P. is $\frac{2a-6b}{3b},\frac{2a-3b}{3b},\frac{2a}{3b},\frac{2a+3b}{3b},\frac{2a+6b}{3b}$....

Common difference (d) = a2 − a1
= $\frac{2a-3b}{3b}-\frac{2a-6b}{3b}$
= $\frac{2a-3b-2a+6b}{3b}$
=​ $\frac{3b}{3b}$
=​ b

Hence, the sequence $\frac{2a-6b}{3b},\frac{2a-3b}{3b},\frac{2a}{3b},\frac{2a+3b}{3b},\frac{2a+6b}{3b}$.... is an A.P. with common difference b.

#### Question 2:

If the sequence $\frac{a-3b}{b},\frac{3a-3b}{b},\frac{5a-3b}{b},\frac{7a-3b}{b},...$is an A.P with common difference 3, then a/b = ________.

The given A.P. is: $\frac{a-3b}{b},\frac{3a-3b}{b},\frac{5a-3b}{b},\frac{7a-3b}{b},...$

Common difference (d) = a2 − a1

Hence, if the sequence $\frac{a-3b}{b},\frac{3a-3b}{b},\frac{5a-3b}{b},\frac{7a-3b}{b},...$is an A.P with common difference 3, then a/b = $\overline{)\frac{3}{2}}$.

#### Question 3:

The 11th term of the A.P.,..., is

The given A.P. is: ,...,

First term (a) = –5
Common difference (d) = a2 – a1
=  $-\frac{5}{2}-\left(-5\right)$
=  $-\frac{5}{2}+5$
=  $\frac{-5+10}{2}$
=  $\frac{5}{2}$

The 11th term is:

​Hence, the 11th term of the A.P.,..., is 20.

#### Question 4:

The 21st term of the A.P. whose first two terms are – 3 and 4 is ____________.

Given:
First term (a) = –3
Second term (a2) = 4
Common difference (d) = a2 – a1
=  $4-\left(-3\right)$
=  4 + 3
=  7

The 21st term is:

​Hence, the 21st term of the A.P. whose first two terms are –3 and 4 is 137.

#### Question 5:

The famous Mathematician associated with finding the sum of the first 100 natural numbers was _________.

Gauss is the famous Mathematician associated with finding the sum of the first 100 natural numbers.

Hence, the famous Mathematician associated with finding the sum of the first 100 natural numbers was Gauss.

#### Question 6:

If the common difference of an A.P. is 5, then a18a13 = ______________.

Given:
Common difference (d) = 5

Now,

Hence, a18 – a13 = 25.

#### Question 7:

If a1, a2, a3, ....., an, ..... is an A.P. such that a18a14 = 32, then its common difference is __________.

Given:
a18 – a14 = 32

Now,

Hence, if a1a2a3, ....., an, ..... is an A.P. such that a18 – a14 = 32, then its common difference is 8.

#### Question 8:

If 5, a2, a3, ...., a20, 145 is an A.P., then a2 + a20 =__________.

If a1a2a3, ...., an are in A.P., then
a+ a= aan−1 aan−= ....

Here,
aa21 aa20
⇒ aa20 = 5 + 145 = 150

Hence, a2 + a20 = 150.

#### Question 9:

If n – 2, 4n – 1, and 5n + 2 are in A.P., then n = _______.

If – 2, 4n – 1, and 5+ 2 are in A.P.,
then their common difference must be same.

Hence, = 1.

#### Question 10:

The value of the middle term of the A.P. –11, –7, –3, ...., 49, 53 is ________.

The given A.P. is: –11, –7, –3, ...., 49, 53

First term (a) = –11
Common difference (d) = a2 – a1
=  –7 – (–11)
=  –7 + 11
=  4
Last term (an) = 53

Now,
${a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}⇒53=-11+\left(n-1\right)\left(4\right)\phantom{\rule{0ex}{0ex}}⇒53=-11+4n-4\phantom{\rule{0ex}{0ex}}⇒53=-15+4n\phantom{\rule{0ex}{0ex}}⇒4n=53+15\phantom{\rule{0ex}{0ex}}⇒4n=68\phantom{\rule{0ex}{0ex}}⇒n=17$

∴ Total number of terms = 17

Therefore,

​Hence, the value of the middle term of the A.P. –11, –7, –3, ...., 49, 53 is 21.

#### Question 11:

If 9th term of an A.P. is zero, then its 29th and 19th terms are in the ratio __________.

Given:
9th term (a9) = 0

Now,
${a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{a}_{9}=a+\left(9-1\right)d\phantom{\rule{0ex}{0ex}}⇒0=a+8d\phantom{\rule{0ex}{0ex}}⇒a=-8d$

Therefore,

​Hence, if 9th term of an A.P. is zero, then its 29th and 19th terms are in the ratio 2 : 1.

#### Question 12:

If an denotes the nth term of the A.P. 3, 8, 13, 18, ..., then the value of a30 a20 is __________.

The given A.P. is: 3, 8, 13, 18, ...

First term  (a) = 3
Common difference (d) = 8 – 3 = 5

Now,

Hence, the value of a30 – a20 is 50.

#### Question 13:

The sum of first 50 odd natural numbers is __________.

First 50 odd natural numbers are: 1, 3, 5, 7, ...

First term (a) = 1
Common difference (d) = 3 – 1 = 2

Now,

Hence, the sum of first 50 odd natural numbers is 2500.

#### Question 14:

The sum of first n odd natural numbers is ________.

First n odd natural numbers are: 1, 3, 5, 7, ...

First term (a) = 1
Common difference (d) = 3 – 1 = 2

Now,

Hence, the sum of first n odd natural numbers is n2.

#### Question 15:

In an A.P. a1, a2, a3, ..., an, ...., if a1 = 1, an = 20 and Sn = 399, then the value of n is __________.

Given:
First term (a) = a1 = 1
nth term = a= 20
Sum of n terms = S= 399

Now,

Hence, the value of is 38.

#### Question 16:

If 7 times the 7th term of an A.P. is equal to 11 times its 11th term, then the value of its 18th term is ________.

Given:
$7{a}_{7}=11{a}_{11}$

Now,

Hence, the value of its 18th term is 0.

#### Question 17:

Two arithmetic progressions have the same common difference. Their first terms are A and B respectively. The difference between their nth terms is _________.

Let the first term of first A.P. be A and common difference be d.
And  the first term of second A.P. be B and common difference be d.

nth term of first A.P. = an
nth term of second A.P. = bn

Now,

Hence, the difference between their nth terms is A − B.

#### Question 18:

If the nth terms of two A.P.s: 9,7,5.... and 24, 21, 18, ... are the same, then the value of n is _________.

The given A.P.s are:
9,7,5.... and 24, 21, 18, ...

9, 7, 5,...
First term (a) = 9
Common difference (d) = 7 − 9 = −2

24, 21, 18, ...
First term (a) = 24
Common difference (d) = 21 − 24 = −3

It is given that, an = bn
$⇒11-2n=27-3n\phantom{\rule{0ex}{0ex}}⇒3n-2n=27-11\phantom{\rule{0ex}{0ex}}⇒n=16$

Hence, the value of is 16.

#### Question 19:

If Sn = n(4n + 1) is the sum of n terms of an A.P., then its common difference is __________.

Given:

Hence, if Sn = n(4n + 1) is the sum of n terms of an A.P., then its common difference is 8.

#### Question 20:

If the ratio of the sums of first n terms of two A.P.s is $\frac{5n+13}{7n+27}$, then the ratio of their 4th terms is ________.

Given:

Hence, the ratio of their 4th terms is 12 : 19.

#### Question 1:

Define an arithmetic progression.

An arithmetic progression is a sequence of terms such that the difference between any two consecutive terms of the sequence is always same.

Suppose we have a sequence

So, if these terms are in A.P., then,

And so on…

Here, d is the common difference of the A.P.

Example: 1, 3, 5, 7, 9 … is an A.P. with common difference (d) as 2.

#### Question 2:

Write the common difference of an A.P. whose nth term is an = 3n + 7.

In the given problem, nth term is given by,

.

To find the common difference of the A.P., we need two consecutive terms of the A.P.

So, let us find the first and the second term of the given A.P.

First term,

Second term (),

Now, the common difference of the A.P. (d) =

Therefore, the common difference is.

#### Question 3:

Which term of the sequence 114, 109, 104, ... is the first negative term?

Here, A.P is

So, first term,

Now,

Common difference (d) =

Now, we need to find the first negative term,

Further simplifying, we get,

Thus,

Therefore, the first negative term is the of the given A.P.

#### Question 4:

Write the value of a30 − a10 for the A.P. 4, 9, 14, 19, ....

In this problem, we are given an A.P. and we need to find.

A.P. is

Here,

First term (a) = 4

Common difference of the A.P. (d)

Now, as we know,

Here, we find a30 and a20.

So, for 30th term,

Also, for 10th term,

So,

Therefore, for the given A.P.

#### Question 5:

Write 5th term from the end of the A.P. 3, 5, 7, 9, ..., 201.

In the given problem, we need to find the 5th term from the end for the given A.P.

3, 5, 7, 9 …201

Here, to find the 5th term from the end let us first find the common difference of the A.P. So,

First term (a) = 3

Last term (an) = 201

Common difference (d) =

Now, as we know, the nth term from the end can be given by the formula,

So, the 5th term from the end,

Therefore, the 5th term from the end of the given A.P. is.

#### Question 6:

Write the value of x for which 2x, x + 10 and 3x + 2 are in A.P.

Here, we are given three terms,

First term (a1) =

Second term (a2) =

Third term (a3) =

We need to find the value of x for which these terms are in A.P. So, in an A.P. the difference of two adjacent terms is always constant. So, we get,

Also,

Now, on equating (1) and (2), we get,

Therefore, for, these three terms will form an A.P.

#### Question 7:

Write the nth term of an A.P. the sum of whose n terms is Sn.

We are given an A.P. the sum of whose n terms is Sn. So, to calculate the nth term of the A.P. we use following formula,

So, the nth term of the A.P. is given by .

#### Question 8:

Write the sum of first n odd natural numbers.

In this problem, we need to find the sum of first n odd natural numbers.

So, we know that the first odd natural number is 1. Also, all the odd terms will form an A.P. with the common difference of 2.

So here,

First term (a) = 1

Common difference (d) = 2

So, let us take the number of terms as n

Now, as we know,

So, for n terms,

Therefore, the sum of first n odd natural numbers is.

#### Question 9:

Write the sum of first n even natural numbers.

In this problem, we need to find the sum of first n even natural numbers.

So, we know that the first odd natural number is 2. Also, all the odd terms will form an A.P. with the common difference of 2.

So here,

First term (a) = 2

Common difference (d) = 2

So, let us take the number of terms as n

Now, as we know,

So, for n terms,

Therefore, the sum of first n even natural numbers is.

#### Question 10:

If the sum of n terms of an A.P. is Sn = 3n2 + 5n. Write its common difference.