#### Page No 9.10:

#### Answer: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

#### Page No 9.10:

#### Answer:2

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

#### Page No 9.11:

#### Answer:3

(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 1^{st} time=

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

Amount left after vacuum pump removes air for 3^{rd} 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.

#### Page No 9.11:

#### Answer:4

In the given problem, we need to show that the given sequence is an A.P and then find its common difference.

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

Substituting *n *= 4, we get

Substituting *n *= 5, we get

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

We find the common difference (*d*)

*d*

Thus,

Also,

Since

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

#### Page No 9.11:

#### Answer:5

In the given problem, we need to show that the given sequence is not an A.P

Here,

Now, first we will find its few terms by substituting

So,

Substituting *n *= 1*, *we get

Substituting *n *= 2*, *we get

Substituting *n *= 3,* *we get

Substituting *n *= 4*, *we get

Substituting *n *= 5*, *we get

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

We find the common difference (*d*)

Thus,

Also,

So,

Hence, the given sequence is not an A.P.

#### Page No 9.11:

#### Answer:6

In the given problem, we need to find that the given sequence is an A.P or not and then find its 15^{th}^{ }term and the common difference.

Here,

Now, to find that it is an A.P or not, 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 sequence to be an A.P,

We find the common difference (*d*)

Thus,

Also,

Since

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

Now, to find its 15^{th} using the formula

First term (*a*) = 11

*n *= 15

Common difference (*d*) = −4

Substituting the above values in the formula

Therefore,

#### Page No 9.11:

#### Answer:7

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 (*a*_{1}) =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 (*a*_{1}) =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 (*a*_{1}) =−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 (*a*_{1}) =−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

#### Page No 9.11:

#### Answer:8

In the given problem, we are given the sequence with the *n*^{th} 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 .

#### Page No 9.11:

#### Answer:9

In the given problem, we are given the sequence with the *n*^{th} term ().

We need to show that these sequences form an A.P

(i)

Now, to show that it is an A.P, we will first 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 sequence to be an A.P,

Common difference (*d*)

Here,

Also,

Since

Hence, the given sequence is an A.P

(ii)

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

(iii)

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

(iv)

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 sequence to be an A.P,

Common difference (*d*)

Here,

Also,

Since

Hence, the given sequence is an A.P.

#### Page No 9.11:

#### Answer:10

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*) = *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,

${a}_{3}-{a}_{2}={7}^{2}-{5}^{2}=49-25=24$

Since

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

#### Page No 9.11:

#### Answer:11

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 5^{th} and 6^{th} 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 5^{th} and 6^{th} 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 5^{th} and 6^{th} 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*) =

^{th} and 6^{th} 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*) =

^{th} and 6^{th} 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.

#### Page No 9.11:

#### Answer:12

In the given problem, *n*th 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.

#### Page No 9.22:

#### Answer:1

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

(i) 10^{th} term of the A.P.

Here,

First term (*a*) = 1

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

Now, as we know,

So, for 10^{th} term,

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

(ii) 18^{th} term of the A.P.

Here,

First term (*a*) =

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

Now, as we know,

So, for 18^{th} term,

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

(iii)* n*^{th} term of the A.P.

Here,

First term (*a*) = 13

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

Now, as we know,

So, for *n*^{th} term,

Therefore, the *n*^{th} term of the given A.P. is.

(iv) 10^{th} term of the A.P.

Here,

First term (*a*) = -40

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

Now, as we know,

So, for 10^{th} term,

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

(v) 8^{th} term of the A.P.

Here,

First term (*a*) = 117

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

Now, as we know,

So, for 8^{th} term,

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

(vi) 11^{th} term of the A.P.

Here,

First term (*a*) = 10.0

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

Now, as we know,

So, for 11^{th} term,

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

(vii) 9^{th} term of the A.P.

Here,

First term (*a*) =

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

Now, as we know,

So, for 9^{th} term,

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

#### Page No 9.22:

#### Answer:2

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.

#### Page No 9.22:

#### Answer:3

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

Therefore the answer is

(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

Therefore the answer is

(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

Therefore, the answer is

#### Page No 9.22:

#### Answer:4

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 .

#### Page No 9.23:

#### Answer:5

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

#### Page No 9.23:

#### Answer:6

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

We need to find the 40^{th} 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,

#### Page No 9.23:

#### Answer:7

In the given problem, the 9^{th} 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 29^{th} term is double of 19^{th} term. So, let us first find the two terms.

For 19^{th} term (*n* = 19),

(Using 1)

For 29^{th} term (*n* = 29),

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

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

Hence proved.

#### Page No 9.23:

#### Answer:8

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 10^{th} term is equal to 15 times the 15^{th} term. We need to show that 25^{th} term is zero.

So, let us first find the two terms.

So, as we know,

For 10^{th} term (*n* = 10),

For 15^{th} term (*n* = 15),

Now, we are given,

Solving this, we get,

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

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

Hence proved

#### Page No 9.23:

#### Answer:9

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

We need to find the 26^{th} 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,

#### Page No 9.23:

#### Answer:10

Here, we are given that 24^{th} term is twice the 10^{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 have to prove that

So, let us first find the two terms.

As we know,

For 10^{th} term (*n* = 10),

For 24^{th} term (*n* = 24),

Now, we are given that

So, we get,

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

For 34^{th} term (*n* = 34),

(Using 1)

For 72^{nd} term (*n* = 72),

(Using 1)

Therefore,

#### Page No 9.23:

#### Answer:11

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 (3*m*+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 (3*m*+1)^{th} term (*n’* = 3*m**+*1),

Therefore,

Hence proved

#### Page No 9.23:

#### Answer:12

Here, we are given two A.P. sequences whose *n*^{th} terms are equal. We need to find *n. *

So let us first find the *n*^{th} 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 *n*^{th} 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 *n*^{th} term,

Now, we are given that the *n*^{th} terms for both the A.P. sequences are equal, we equate (1) and (2),

Therefore,

#### Page No 9.23:

#### Answer:13

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

(i) 3, 5, 7, 9 …201

Here, to find the 12^{th} 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 (*a*_{n}) = 201

Common difference (*d*) =

Now, as we know,

So, for the last term,

Further simplifying,

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

So, for the 89^{th} term (*n =* 89)

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

(ii) 3, 8, 13 …253

Here, to find the 12^{th} 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 (*a*_{n}) = 253

Common difference, *d* =

Now, as we know,

So, for the last term,

Further simplifying,

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

So, for the 40^{th} term (*n =* 40)

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

(iii) 1, 4, 7, 10 …88

Here, to find the 12^{th} 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 (*a*_{n}) = 88

Common difference, *d*= =3

Now, as we know,

So, for the last term,

Further simplifying,

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

So, for the 19^{th} term (*n =* 19)

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

#### Page No 9.23:

#### Answer:14

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 4^{th} term (*n = *4),

Similarly, for the 3^{rd} term (*n = *3),

Also, for the 7^{th} term (*n = *7),

Now, using the value of *a*_{3} 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

#### Page No 9.23:

#### Answer:15

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

We need to find the 2^{nd} and *n*^{th} 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 2^{nd} term (*n *= 2),

For the *n*^{th} term,

Therefore,

#### Page No 9.23:

#### Answer:16

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 (*a*_{n}) = 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.

#### Page No 9.23:

#### Answer:17

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

Here we are given the following,

First term (*a*) = 7

Last term (*a*_{n}) = 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 32^{nd} term (*n =* 32)

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

#### Page No 9.23:

#### Answer:18

In the given problem, the sum of 4^{th} and 8^{th} term is 24 and the sum of 6^{th} and 10^{th} 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 4^{th} term (*n* = 4),

For 8^{th} term (*n* = 8),

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

Also, for 6^{th} term (*n* = 6),

For 10^{th} 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

#### Page No 9.23:

#### Answer:19

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

We need to find the 50^{th} 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,

#### Page No 9.23:

#### Answer:20

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 30^{th} term,

Also, for 20^{th} 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 *a*_{30} and *a*_{20}*.*

So, for 30^{th} term,

Also, for 20^{th} term,

So,

Therefore, for the given A.P

#### Page No 9.23:

#### Answer:21

A.P: *a, a+d, a+2d …*

Here, we first need to write the expression for

Now, as we know,

So, for *n*^{th} term,

Similarly, for *k*^{th} term

So,

So,

(i) In the given problem, we are given 11^{th} and 13^{th}^{ }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 *a*_{30} and *a*_{20}*.*

So, for 10^{th} term,

Also, for 5^{th} term,

So,

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

(iii) In the given problem, the 20^{th} term is 10 more than the 18^{th} term. So, let us first find the 20^{th} term and 18^{th} 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 20^{th} term (*n* = 20),

Also, for 18^{th} term (*n* = 18),

Now, we are given,

On substituting the values, we get,

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

#### Page No 9.23:

#### Answer:22

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 (*a*_{n}) = 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 (*a*_{n}) = −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 (*a*_{n}) = 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 (*a*_{n}) =

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.

#### Page No 9.23:

#### Answer:23

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

Here,

The first term (*a*) = *a*

The last term (*a*_{n}) = *l*

Now, as we know,

So, for the *m*^{th}^{ }term from the beginning, we take (*n = m*),

Similarly, for the *m*^{th} 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

#### Page No 9.23:

#### Answer:24

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 3^{rd} term (*n* = 3),

Also, for 5^{th} term (*n* = 5),

For 7^{th} 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

#### Page No 9.24:

#### Answer:37

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

We know that, *n*^{th }term = *a*_{n }= *a* + (*n* − 1)*d*

According to the question,

*a*_{17} = 5 + 2*a*_{8}

⇒ *a* + (17 − 1)*d* = 5 + 2(*a* + (8 − 1)*d*)

⇒ *a* + 16*d* = 5 + 2*a* + 14*d*

⇒ 16*d* − 14*d *= 5 + 2*a* − *a*

⇒ 2*d *= 5 + *a*

⇒ *a = *2*d *− 5 * ....* (1)

Also, *a*_{11} = 43

⇒ *a* + (11 − 1)*d* = 43

⇒ *a* + 10*d* = 43 ....(2)

On substituting the values of (1) in (2), we get

2*d *− 5 + 10*d* = 43

⇒ 12*d *= 5 + 43

⇒ 12*d *= 48

⇒ *d *= 4

⇒ *a* = 2 × 4 − 5 [From (1)]

⇒ *a* = 3

∴ *a*_{n }= *a* + (*n* − 1)*d
=* 3 + (

*n*− 1)4

= 3 + 4

*n*− 4

= 4

*n*− 1

Thus, the

*n*

^{th}term of the given A.P. is 4

*n*− 1.

#### Page No 9.24:

#### Answer:38

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, *n*^{th }term = *a*_{n }= *a* + (*n* − 1)*d*

According to the question,

999 = 108 + (*n* − 1)9

⇒ 108 + 9*n* − 9 = 999

⇒ 99 + 9*n* = 999

⇒ 9*n* = 999 − 99

⇒ 9*n* = 900

⇒ *n* = 100

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

#### Page No 9.24:

#### Answer:39

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

We know that, *n*^{th }term = *a*_{n }= *a* + (*n* − 1)*d*

According to the question,

*a*_{19} = 3*a*_{6}

⇒ *a* + (19 − 1)*d* = 3(*a* + (6 − 1)*d*)

⇒ *a* + 18*d* = 3*a* + 15*d*

⇒ 18*d* − 15*d *= 3*a* − *a*

⇒ 3*d *= 2*a*

⇒ *a = $\frac{3}{2}$**d* * ....* (1)

Also, *a*_{9} = 19

⇒ *a* + (9 − 1)*d* = 19

⇒ *a* + 8*d* = 19 ....(2)

On substituting the values of (1) in (2), we get

*$\frac{3}{2}$d *+ 8*d* = 19

⇒ 3*d *+ 16*d *= 19 × 2

⇒ 19*d *= 38

⇒ *d *= 2

⇒ *a* = $\frac{3}{2}\times 2$ [From (1)]

⇒ *a* = 3

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

#### Page No 9.24:

#### Answer:40

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

We know that, *n*^{th }term = *a*_{n }= *a* + (*n* − 1)*d*

According to the question,

*a*_{9} = 6*a*_{2}

⇒ *a* + (9 − 1)*d* = 6(*a* + (2 − 1)*d*)

⇒ *a* + 8*d* = 6*a* + 6*d*

⇒ 8*d* − 6*d *= 6*a* − *a*

⇒ 2*d *= 5*a*

⇒ *a = $\frac{2}{5}$**d* * ....* (1)

Also, *a*_{5} = 22

⇒ *a* + (5 − 1)*d* = 22

⇒ *a* + 4*d* = 22 ....(2)

On substituting the values of (1) in (2), we get

*$\frac{2}{5}$d *+ 4*d* = 22

⇒ 2*d *+ 20*d *= 22 × 5

⇒ 22*d *= 110

⇒ *d *= 5

⇒ *a* = $\frac{2}{5}\times 5$ [From (1)]

⇒ *a* = 2

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

#### Page No 9.24:

#### Answer:41

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

We know that, *n*^{th }term = *a*_{n }= *a* + (*n* − 1)*d*

According to the question,

*a*_{24} = 2*a*_{10}

⇒ *a* + (24 − 1)*d* = 2(*a* + (10 − 1)*d*)

⇒ *a* + 23*d* = 2*a* + 18*d*

⇒ 23*d* − 18*d *= 2*a* − *a*

⇒ 5*d *= *a*

⇒ *a = *5*d* * ....* (1)

Also,

*a*_{72} = *a* + (72 − 1)*d*

= 5*d* + 71*d * [From (1)]

= 76*d * ..... (2)

and

*a*_{15 }= *a* + (15 − 1)*d*

= 5*d* + 14*d * [From (1)]

= 19*d * ..... (3)

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

76*d *= 4 × 19*d*

⇒ *a*_{72} = 4 × *a*_{15}

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

#### Page No 9.24:

#### Answer:25

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 7^{th} term (*n* = 7),

Also, for 13^{th} 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

#### Page No 9.24:

#### Answer:26

In the given problem, let us first find the 13^{th} 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 13^{th} term (*n* = 13),

Let us take the term which is 84 more than the 13^{th} term as *a*_{n}. So,

Also,

Further simplifying, we get,

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

#### Page No 9.24:

#### Answer:27

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 100^{th} terms is 100.

We need to find the difference between their 100^{th} terms

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

Now, as we know,

So, for 100^{th} term of first A.P. (*n* = 100),

Now, for 100^{th} term of second A.P. (*n* = 100),

Now, we are given,

On substituting the values, we get,

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

So, for 1000^{th} term of first A.P. (*n* = 1000),

Now, for 1000^{th} term of second A.P. (*n* = 1000),

So,

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

#### Page No 9.24:

#### Answer:28

Here, we are given two A.P. sequences. We need to find the value of *n* for which the *n*^{th }terms of both the sequences are equal. We need to find *n*

So let us first find the *n*^{th} 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 *n*^{th} 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 *n*^{th} term,

Now, we are given that the *n*^{th} terms for both the A.P. sequences are equal, we equate (1) and (2),

Therefore,

#### Page No 9.24:

#### Answer:29

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 (*a*_{n}) = 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.

#### Page No 9.24:

#### Answer:30

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 (*a*_{n}) = 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.

#### Page No 9.24:

#### Answer:31

In the given problem, let us first find the 41^{st} 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 41^{st} term (*n* = 41),

Let us take the term which is 72 more than the 41^{st} term as *a*_{n}. So,

Also,

Further simplifying, we get,

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

#### Page No 9.24:

#### Answer:32

In the given problem, let us first find the 36^{st} 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 36^{th} term (*n* = 36),

Let us take the term which is 39 more than the 36^{th} term as *a*_{n}. So,

Also,

Further simplifying, we get,

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

#### Page No 9.24:

#### Answer:33

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

We have the A.P as 7, 10, 13 …184

Here, to find the 8^{th} 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 (*a*_{n}) = 184

Common difference (*d*) = =3

Now, as we know,

So, for the last term,

Further simplifying,

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

So, for the 53^{rd} term (*n =* 53)

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

#### Page No 9.24:

#### Answer:34

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

We have the A.P as 8, 10, 12 …126

Here, to find the 10^{th} 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 (*a*_{n}) = 126

Common difference (*d*) = =2

Now, as we know,

So, for the last term,

Further simplifying,

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

So, for the 51^{st} term (*n =* 51)

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

#### Page No 9.24:

#### Answer:35

In the given problem, the sum of 4^{th} and 8^{th} term is 24 and the sum of 6^{th} and 10^{th} 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 4^{th} term (*n* = 4),

For 8^{th} term (*n* = 8),

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

Also, for 6^{th} term (*n* = 6),

For 10^{th} 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.

#### Page No 9.24:

#### Answer:36

In the given problem, let us first find the 21^{st} 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 21^{st} term (*n* = 21),

Let us take the term which is 120 more than the 21^{st} term as *a*_{n}. So,

Also,

Further simplifying, we get,

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

#### Page No 9.27:

#### Answer:1

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 .

#### Page No 9.27:

#### Answer:2

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 .

#### Page No 9.27:

#### Answer:3

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 .

#### Page No 9.27:

#### Answer:4

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 .

#### Page No 9.27:

#### Answer:5

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 .

#### Page No 9.27:

#### Answer:6

Here, we are given three terms,

First term (*a*_{1}) =

Second term (*a*_{2}) =

Third term (*a*_{3}) =

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.

#### Page No 9.27:

#### Answer:7

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

First term (*a*_{1}) =

Second term (*a*_{2}) =

Third term (*a*_{3}) =

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.

#### Page No 9.27:

#### Answer:8

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

First term (*a*_{1}) =

Second term (*a*_{2}) =

Third term (*a*_{3}) =

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 2*ab* as the common difference.

Hence proved

#### Page No 9.46:

#### Answer:1

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.

#### Page No 9.46:

#### Answer:2

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.

#### Page No 9.46:

#### Answer:3

Here, we are given an A.P., whose *n*^{th} 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 *n*^{th} term of A.P.

Now, the last term (*l*) or the *n*^{th} 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.

#### Page No 9.46:

#### Answer:4

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,

$3{n}^{2}-53n+232=0$

On solving by splitting the middle term, we get,

$3{n}^{2}-24n-29n+232=0\phantom{\rule{0ex}{0ex}}3n\left(n-8\right)-29\left(n-8\right)=0\phantom{\rule{0ex}{0ex}}\left(3n-29\right)\left(n-8\right)=0$

Further,

Also,

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

So, the term is *a*_{8}

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

#### Page No 9.46:

#### Answer:5

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.

*n *= number of terms

The first term (*a*) = 18

The sum of* n* terms (*S*_{n}) = 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 (*a*_{1}) = −14

Fifth term (*a*_{5}) = 2

Sum of terms (*s*_{n}) = 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.

*n *= number of terms

The first term (*a*_{1}) = −14

The sum of* n* terms (*S*_{n}) = 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.

*n *= number of terms

The first term (*a*) = 9

The sum of* n* terms (*S*_{n}) = 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.

*n *= number of terms

The first term (*a*) = 63

The sum of* n* terms (*S*_{n}) = 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, 22^{nd} 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 = $\frac{x}{2}\left[2\times 27+\left(x-1\right)\left(-3\right)\right]=0$

$\Rightarrow \frac{x}{2}\left[54+\left(-3x+3\right)\right]=0\phantom{\rule{0ex}{0ex}}\Rightarrow x\left(54-3x+3\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x\left(57-3x\right)=0$

Now, either *x* = 0 or 57 − 3*x* = 0.

Since the number of terms cannot be 0, $x\ne 0$.

∴ 57 − 3*x* = 0

⇒ 57 = 3*x*

⇒ *x* = 19

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

#### Page No 9.47:

#### Answer:6

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.

#### Page No 9.47:

#### Answer:7

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 3^{th} term (*n = *3),

Similarly, for the 7^{th} 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 .

#### Page No 9.47:

#### Answer:8

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.

#### Page No 9.47:

#### Answer:9

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 12^{th} 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 .

#### Page No 9.47:

#### Answer:10

In the given problem, we need to find the sum of first 22 terms of an A.P. Let us take the first term as *a*.

Here, we are given that,

Also, we know,

For the 22^{nd} term (*n = *22),

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* = 22, we get,

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

#### Page No 9.47:

#### Answer:11

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

#### Page No 9.47:

#### Answer:12

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.

#### Page No 9.47:

#### Answer:13

(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.

#### Page No 9.47:

#### Answer:14

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

#### Page No 9.47:

#### Answer:15

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.

#### Page No 9.47:

#### Answer:16

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.

#### Page No 9.47:

#### Answer:17

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.

#### Page No 9.47:

#### Answer:18

(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

$=\left[\left(1+2+.....+549\right)-\left(1+2+.....100\right)\right]-16425\phantom{\rule{0ex}{0ex}}=\left[\left(\frac{549\times 550}{2}\right)-\left(\frac{100\times 101}{2}\right)\right]-16425\phantom{\rule{0ex}{0ex}}=150975-5050-16425\phantom{\rule{0ex}{0ex}}=129500\phantom{\rule{0ex}{0ex}}$

(iii)

Integers between 1 and 500 which are multiples of both 2 and 5 are

10, 20, 30, ....., 490.

${a}_{n}=490\phantom{\rule{0ex}{0ex}}\Rightarrow 10+\left(n-1\right)10=490\phantom{\rule{0ex}{0ex}}\Rightarrow \left(n-1\right)10=480\phantom{\rule{0ex}{0ex}}\Rightarrow n=49\phantom{\rule{0ex}{0ex}}Now,\phantom{\rule{0ex}{0ex}}{S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}{S}_{49}=\frac{49}{2}\left[20+\left(48\right)10\right]\phantom{\rule{0ex}{0ex}}{S}_{49}=\frac{49}{2}\left[500\right]\phantom{\rule{0ex}{0ex}}{S}_{49}=49\times 250=12250$

(iv)

Integers from 1 and 500 which are multiples of both 2 and 5 are

10, 20, 30, ....., 500.

${a}_{n}=500\phantom{\rule{0ex}{0ex}}\Rightarrow 10+\left(n-1\right)10=500\phantom{\rule{0ex}{0ex}}\Rightarrow \left(n-1\right)10=490\phantom{\rule{0ex}{0ex}}\Rightarrow n=50\phantom{\rule{0ex}{0ex}}Now,\phantom{\rule{0ex}{0ex}}{S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}{S}_{50}=\frac{50}{2}\left[20+\left(49\right)10\right]\phantom{\rule{0ex}{0ex}}=25\left[510\right]\phantom{\rule{0ex}{0ex}}=12750$

(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

${a}_{n}=500\phantom{\rule{0ex}{0ex}}\Rightarrow 2+\left(n-1\right)2=500\phantom{\rule{0ex}{0ex}}\Rightarrow n=250\phantom{\rule{0ex}{0ex}}{S}_{250}=\frac{250}{2}\left(4+249\times 2\right)=125\times 502=62750\phantom{\rule{0ex}{0ex}}$

and

Multiples of 5 are 5, 10, 15, ....., 500. Therefore sum of all the multiples of 5 from 1 and 500 is

${a}_{n}=500\phantom{\rule{0ex}{0ex}}\Rightarrow 5+\left(n-1\right)5=500\phantom{\rule{0ex}{0ex}}\Rightarrow n=100\phantom{\rule{0ex}{0ex}}{S}_{100}=\frac{100}{2}\left(10+99\times 5\right)=50\times 505=25250\phantom{\rule{0ex}{0ex}}$

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

${a}_{n}=500\phantom{\rule{0ex}{0ex}}\Rightarrow 10+\left(n-1\right)10=500\phantom{\rule{0ex}{0ex}}\Rightarrow \left(n-1\right)10=490\phantom{\rule{0ex}{0ex}}\Rightarrow n=50\phantom{\rule{0ex}{0ex}}Now,\phantom{\rule{0ex}{0ex}}{S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}{S}_{50}=\frac{50}{2}\left[20+\left(49\right)10\right]\phantom{\rule{0ex}{0ex}}=25\left[510\right]\phantom{\rule{0ex}{0ex}}=12750$

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

#### Page No 9.47:

#### Answer:19

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,

#### Page No 9.47:

#### Answer:20

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 5^{th} term (*n = *5),

Similarly, for the 12^{th} 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.

#### Page No 9.47:

#### Answer:21

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,

${a}_{4}=a+3d\phantom{\rule{0ex}{0ex}}8=a+3d...\left(2\right)$

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.

#### Page No 9.47:

#### Answer:22

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,

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

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

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

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

#### Page No 9.47:

#### Answer:23

*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

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

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

#### Page No 9.47:

#### Answer:24

(i) Here, we are given an A.P. whose *n*^{th} 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 *n*^{th} term of A.P.

Now, the last term (*l*) or the *n*^{th} 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 *n*^{th} 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 *n*^{th} term of A.P.

Now, the last term (*l*) or the *n*^{th} 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 *n*^{th} 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 *n*^{th} term of A.P.

Now, the last term (*l*) or the *n*^{th} 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 *n*^{th} 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 *n*^{th} term of A.P.

Now, the last term (*l*) or the *n*^{th} 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.

#### Page No 9.48:

#### Answer:25

Here, we are given an A.P. whose *n*^{th} 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 *n*^{th} term of A.P.

Now, the last term (*l*) or the *n*^{th} 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.

#### Page No 9.48:

#### Answer:26

Here, we are given an A.P. whose *n*^{th} 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 *n*^{th} term of A.P.

Now, the last term (*l*) or the *n*^{th} 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.

#### Page No 9.48:

#### Answer:27

Here, we are given an A.P. whose *n*^{th} 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 *n*^{th} term of A.P.

Now, the last term (*l*) or the *n*^{th} 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.

#### Page No 9.48:

#### Answer:28

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 2^{nd} term (*n = *2),

Similarly, for the 3^{rd} 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.

#### Page No 9.48:

#### Answer:29

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.

*n *= 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.

#### Page No 9.48:

#### Answer:30

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.

#### Page No 9.48:

#### Answer:31

Here, the sum of first *n* terms is given by the expression,

We need to find the 25^{th} term of the A.P.

So we know that the *n*^{th}term 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 (*S*_{25}) and the sum of 24 terms (*S*_{24}). We get,

Similarly,

Now, using the above values in (1),

Therefore, .

#### Page No 9.48:

#### Answer:32

(i) Here, we have an A.P. whose *n*^{th} term (*a*_{n}), 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 (*S*_{n}).

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 *n*^{th} term (*a*_{n}), sum of first *n* terms (*S*_{n}) 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 (*S*_{n}) = −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 (*S*_{n}) and the number of terms (*n*) are given. We need to find common difference (*d*).

Here,

First term () = 3

Sum of *n* terms (*S*_{n}) = 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 *n*^{th} term (*a*_{n}), sum of first *n* terms (*S*_{n}) and the number of terms (*n*) are given. We need to find first term (*a*).

Here,

Last term () = 28

Sum of *n* terms (*S*_{n}) = 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 *n*^{th} term (*a*_{n}), sum of first *n* terms (*S*_{n}) 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 (*S*_{n}) = 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 *n*^{th} term (*a*_{n}).

Here,

First term (*a*) = 2

Sum of first *n*^{th} 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 *n*^{th} term (*a*_{n}) using the formula,

So, substituting the values in the above mentioned formula

Therefore, for the given A.P.

(vii)

${a}_{k}={S}_{k}-{S}_{k-1}\phantom{\rule{0ex}{0ex}}\Rightarrow 164=\left(3{k}^{2}+5k\right)-\left(3{\left(k-1\right)}^{2}+5\left(k-1\right)\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 164=3{k}^{2}+5k-3{k}^{2}+6k-3-5k+5\phantom{\rule{0ex}{0ex}}\Rightarrow 164=6k+2\phantom{\rule{0ex}{0ex}}\Rightarrow 6k=162\phantom{\rule{0ex}{0ex}}\Rightarrow k=27\phantom{\rule{0ex}{0ex}}$

#### Page No 9.48:

#### Answer:33

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.

#### Page No 9.48:

#### Answer:34

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,

We get a quadratic equation,

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.

#### Page No 9.48:

#### Answer:35

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.

#### Page No 9.48:

#### Answer:36

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.

#### Page No 9.48:

#### Answer:37

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 31^{st} minute the rate of counting per minute = 177.

At 32^{nd} 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

$\mathrm{Now}\mathrm{let}\mathrm{n}=60\mathrm{then}\mathrm{finding}\mathrm{the}\mathrm{last}\mathrm{term},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[\mathrm{a}+\mathrm{l}\right]\phantom{\rule{0ex}{0ex}}5310=\frac{60}{2}\left[177+\mathrm{l}\right]\phantom{\rule{0ex}{0ex}}177=177+\mathrm{l}\phantom{\rule{0ex}{0ex}}\mathrm{l}=0\phantom{\rule{0ex}{0ex}}\mathrm{It}\mathrm{means}\mathrm{the}\mathrm{work}\mathrm{will}\mathrm{be}\mathrm{finesh}\mathrm{in}59\mathrm{th}\mathrm{minute}\mathrm{only}\mathrm{because}60\mathrm{th}\mathrm{term}\mathrm{is}0.\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{we}\mathrm{will}\mathrm{take}\mathrm{n}=59$

Therefore, the total time required for counting the entire amount

So, the total time required for counting the entire amount is.

#### Page No 9.48:

#### Answer:38

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.

#### Page No 9.48:

#### Answer:39

In the given problem,

Total amount of money (*S*_{n}) = 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

#### Page No 9.48:

#### Answer:40

In the given problem, we have the first and the *n*th 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 *n*th 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..

#### Page No 9.48:

#### Answer:41

In the given problem, we have the first and the *n*th 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 *n*th 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..

#### Page No 9.49:

#### Answer:46

First term, *a* = 10

Sum of first 14 terms, ${S}_{14}=1505$

$\Rightarrow \frac{14}{2}\left[2\times 10+\left(14-1\right)d\right]=1505\phantom{\rule{0ex}{0ex}}\Rightarrow 7\times \left(20-13d\right)=1505\phantom{\rule{0ex}{0ex}}\Rightarrow 20-13d=\frac{1505}{7}=215\phantom{\rule{0ex}{0ex}}\Rightarrow 13d=-195\phantom{\rule{0ex}{0ex}}\Rightarrow d=-15$

Now,

${a}_{25}=10+24\left(-15\right)=-350$

#### Page No 9.49:

#### Answer:47

${S}_{n}=5{n}^{2}+3n\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{know}\phantom{\rule{0ex}{0ex}}{a}_{n}={S}_{n}-{S}_{n-1}\phantom{\rule{0ex}{0ex}}\therefore {a}_{n}=5{n}^{2}+3n-5{\left(n-1\right)}^{2}-3\left(n-1\right)\phantom{\rule{0ex}{0ex}}{a}_{n}=10n-2$

Now,

${a}_{m}=168\phantom{\rule{0ex}{0ex}}\Rightarrow 10m-2=168\phantom{\rule{0ex}{0ex}}\Rightarrow 10m=170\phantom{\rule{0ex}{0ex}}\Rightarrow m=17$

${a}_{20}=10\left(20\right)-2=198$

#### Page No 9.49:

#### Answer:48

${S}_{q}=63q-3{q}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{know}\phantom{\rule{0ex}{0ex}}{a}_{q}={S}_{q}-{S}_{q-1}\phantom{\rule{0ex}{0ex}}\therefore {a}_{q}=63q-3{q}^{2}-63\left(q-1\right)+3{\left(q-1\right)}^{2}\phantom{\rule{0ex}{0ex}}{a}_{q}=66-6q\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},{a}_{p}=-60\phantom{\rule{0ex}{0ex}}\Rightarrow 66-6p=-60\phantom{\rule{0ex}{0ex}}\Rightarrow 126=6p\phantom{\rule{0ex}{0ex}}\Rightarrow p=21\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{a}_{11}=66-6\times 11=0$

#### Page No 9.49:

#### Answer:49

${S}_{m}=4{m}^{2}-m\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{know}\phantom{\rule{0ex}{0ex}}{a}_{m}={S}_{m}-{S}_{m-1}\phantom{\rule{0ex}{0ex}}\therefore {a}_{m}=4{m}^{2}-m-4{\left(m-1\right)}^{2}+\left(m-1\right)\phantom{\rule{0ex}{0ex}}{a}_{m}=8m-5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}{a}_{n}=107\phantom{\rule{0ex}{0ex}}\Rightarrow 8n-5=107\phantom{\rule{0ex}{0ex}}\Rightarrow 8n=112\phantom{\rule{0ex}{0ex}}\Rightarrow n=14\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{a}_{21}=8\left(21\right)-5=163$

#### Page No 9.49:

#### Answer:50

${a}_{n}=-4n+15\phantom{\rule{0ex}{0ex}}\Rightarrow {a}_{1}=-4+15=11\phantom{\rule{0ex}{0ex}}\mathrm{Also},{a}_{2}=-8+15=7\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Common}\mathrm{difference},d={a}_{2}-{a}_{1}=7-11=-4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}{S}_{20}=\frac{20}{2}\left[2\times 11+\left(20-1\right)\left(-4\right)\right]\phantom{\rule{0ex}{0ex}}=10\left(22-76\right)\phantom{\rule{0ex}{0ex}}=-540$

#### Page No 9.49:

#### Answer:51

First term, ${a}_{1}=-12$

Common difference, $d={a}_{2}-{a}_{1}=-9-\left(-12\right)=3$

${a}_{n}=21\phantom{\rule{0ex}{0ex}}\Rightarrow a+\left(n-1\right)d=21\phantom{\rule{0ex}{0ex}}\Rightarrow -12+\left(n-1\right)\times 3=21\phantom{\rule{0ex}{0ex}}\Rightarrow 3n=36\phantom{\rule{0ex}{0ex}}\Rightarrow n=12$

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

$={S}_{12}+12\phantom{\rule{0ex}{0ex}}=\frac{12}{2}\left[2\left(-12\right)+11\left(3\right)\right]+12\phantom{\rule{0ex}{0ex}}=6\times \left(9\right)+12\phantom{\rule{0ex}{0ex}}=66\phantom{\rule{0ex}{0ex}}$

#### Page No 9.49:

#### Answer:52

${S}_{n}=3{n}^{2}+4n$

We know

${a}_{n}={S}_{n}-{S}_{n-1}\phantom{\rule{0ex}{0ex}}\therefore {a}_{n}=3{n}^{2}+4n-3{\left(n-1\right)}^{2}-4\left(n-1\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {a}_{n}=6n+1$

${a}_{25}=6\left(25\right)+1=151$

#### Page No 9.49:

#### Answer:42

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 *n*^{th} term of the A.P.

So, for the third term,

Also, for the tenth term,

So, for the *n*^{th} term,

Therefore, .

#### Page No 9.49:

#### Answer:43

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.

#### Page No 9.49:

#### Answer:44

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 (*S*_{n}) = 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.

#### Page No 9.49:

#### Answer:45

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 *a*_{10}. For the sum of first 10 terms of this A.P,

First term = *a*

Last term = *a*_{10}

So, we know,

For the 10^{th} 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 (*S*_{10}),

First term = *a*_{11}

Last term = *a*_{20}

For the 11^{th} term (*n = *11),

For the 20^{th} term (*n = *20),

So, for the given A.P,

Now, subtracting (1) from (2),

Substituting the value of *d *in (1)

So, the A.P. is with .

#### Page No 9.50:

#### Answer:1

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.

#### Page No 9.50:

#### Answer:2

In the given problem, *n*th 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.

#### Page No 9.50:

#### Answer:3

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.

#### Page No 9.50:

#### Answer:4

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 *a*_{30} and *a*_{20}*.*

So, for 30^{th} term,

Also, for 10^{th} term,

So,

Therefore, for the given A.P.

#### Page No 9.50:

#### Answer:5

In the given problem, we need to find the 5^{th} term from the end for the given A.P.

3, 5, 7, 9 …201

Here, to find the 5^{th} term from the end let us first find the common difference of the A.P. So,

First term (*a*) = 3

Last term (*a*_{n}) = 201

Common difference (*d*) =

Now, as we know, the *n*^{th} term from the end can be given by the formula,

So, the 5^{th} term from the end,

Therefore, the 5^{th} term from the end of the given A.P. is.

#### Page No 9.50:

#### Answer:6

Here, we are given three terms,

First term (*a*_{1}) =

Second term (*a*_{2}) =

Third term (*a*_{3}) =

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.

#### Page No 9.50:

#### Answer:7

We are given an A.P. the sum of whose *n* terms is *S*_{n}. So, to calculate the *n*^{th} term of the A.P. we use following formula,

So, the* n*^{th} term of the A.P. is given by .

#### Page No 9.50:

#### Answer:8

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.

#### Page No 9.50:

#### Answer:9

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.

#### Page No 9.50:

#### Answer:10

Here, we are given,

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

Now, as we know,

So, we get,

Also,

On comparing the terms containing *n *in (1) and (2), we get,

Therefore, the common difference is.

#### Page No 9.50:

#### Answer:11

Here, we are given

First term = *a*

Last term = *b*

Let us take the common difference as *d*

Now, we know

So,

For the last term (*a*_{n}),

Therefore, common difference of the A.P. is

#### Page No 9.50:

#### Answer:12

Here, we are given,

First term (*a*) = *p*

Common difference (*d*) = *q*

We need to find the 10^{th} term (*a*_{n}).

As we know,

So, for 10^{th} term (*n *= 10), we get,

Therefore,

#### Page No 9.50:

#### Answer:13

Here, we are given three terms,

First term (*a*_{1}) =

Second term (*a*_{2}) =

Third term (*a*_{3}) =

We need to find the value of *p* 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.

#### Page No 9.50:

#### Answer:14

Here, we are given three consecutive terms of an A.P.

First term (*a*_{1}) =

Second term (*a*_{2}) =

Third term (*a*_{3}) =

We need to find the value of *a.* 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,

#### Page No 9.50:

#### Answer:15

Here, we are given,

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

Now, as we know,

So, we get,

Also,

On comparing the terms containing *p *in (1) and (2), we get,

Therefore, the common difference is

#### Page No 9.5:

#### Answer:1

Here, we are given the *n*^{th }term for various sequences. We need to find the first five terms of the sequence.

(i)

Here, the *n*^{th} 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 *n*^{th} 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 *n*^{th} 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 *n*^{th} 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 *n*^{th} 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 *n*^{th} 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 *n*^{th} 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 *n*^{th} 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 *n*^{th} 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

#### Page No 9.51:

#### Answer:1

In the given problem, we are given 7^{th} and 13^{th}^{ }term of an A.P.

We need to find the 26^{th} 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 18^{th} term (*n = *18),

Substituting the above values in the formula,

Therefore,

Hence, the correct option is **(c).**

#### Page No 9.51:

#### Answer:2

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).

#### Page No 9.51:

#### Answer:3

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)**.

#### Page No 9.51:

#### Answer:4

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 (*a*_{n}) = 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)**.

#### Page No 9.51:

#### Answer:5

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 *n*^{th} term

So we know that the *n*^{th}term 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).**

#### Page No 9.51:

#### Answer:6

Here, the sum of first *n* terms is given by the expression,

We need to find the *n*^{th} term.

So we know that the *n*^{th}term of an A.P. is given by,

So,

Using the property,

We get,

Therefore,

Hence the correct option is **(c).**

#### Page No 9.51:

#### Answer:7

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)**.

#### Page No 9.51:

#### Answer:8

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)**.

#### Page No 9.51:

#### Answer:9

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 = *S*_{n}

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)**.

#### Page No 9.51:

#### Answer:10

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).**

#### Page No 9.51:

#### Answer:11

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)**.

#### Page No 9.51:

#### Answer:12

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 (*a*_{2}) = b

Last term (*l*__)__ = 2*a*

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).**

#### Page No 9.51:

#### Answer:13

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 *a*_{1},* **a*_{2}….* a*_{n}* *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 = 2*d*

So,

Now,

Thus,

Therefore, the correct option is **(a)**.

#### Page No 9.52:

#### Answer:14

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,

$\frac{1}{2n}\left(2a+nd-d\right)=\frac{1}{2m}\left(2a+md-d\right)$

$\Rightarrow m\left(2a+nd-d\right)=n\left(2a+md-d\right)$

$\Rightarrow 2am+mnd-md=2an+mnd-nd$

Solving further, we get,

Further, substituting (3) in (1), we get,

Now,

Thus,

Therefore, the correct option is **(C).**

#### Page No 9.52:

#### Answer:15

Here, we are given an A.P. whose sum of *n* terms is *S*_{n} 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 *S*_{3n},

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)**.

#### Page No 9.52:

#### Answer:16

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).**

#### Page No 9.52:

#### Answer:17

Here, we are given an A.P. whose sum of *r* terms is *S*_{r}. 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 *S*_{3n},

Similarly,

Also,

So, using (1), (2) and (3), we get,

Taking common, we get,

Therefore,

Hence, the correct option is **(c)**.

#### Page No 9.52:

#### Answer:18

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)**.

#### Page No 9.52:

#### Answer:19

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 (*S*_{n}) = 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)**.

#### Page No 9.52:

#### Answer:20

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).**

#### Page No 9.52:

#### Answer:21

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 9^{th} term (*n = *9),

Similarly, for the 449^{th} 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 *n*^{th} term. So,

So,

Therefore, the correct option is **(d).**

#### Page No 9.52:

#### Answer:22

Here, we are given three terms,

First term (*a*_{1}) =

Second term (*a*_{2}) =

Third term (*a*_{3}) =

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)**.

#### Page No 9.52:

#### Answer:23

A.P. we use following formula,

So, the* n*^{th} term of the A.P. is given by . Therefore, the correct option is **(b).**

#### Page No 9.52:

#### Answer:24

Here, we are given an A.P. the sum of whose *n* terms is *S*_{n}. So, to calculate the common difference of the A.P, we find two consecutive terms of the A.P.

Now, the *n*^{th} 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).**

#### Page No 9.52:

#### Answer:25

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 *n*^{th }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 *n*^{th}* *term, we replace *n* by. We get,

As we know,

Therefore, we get,

Hence the correct option is **(b)**.

#### Page No 9.52:

#### Answer:26

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 *S*_{n}*.*

We need to find the relation between *a* and *d* such thatis independent of

So, let us first find the values of *S*_{x}* *and *S*_{kx} 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)**.

#### Page No 9.53:

#### Answer:36

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

We know that, sum of first* n*^{ }terms = *S*_{n }= $\frac{n}{2}$[2*a* + (*n* − 1)*d*]

The given series is 1 + 3 + 5 + ......

First term = *a *= 1.

Common difference = *d* = 3 − 1 = 2

∴ *S*_{20 }= $\frac{20}{2}$[2 × 1 + (20 − 1)2]

= 10(2 + 19 × 2)

= 10(40)

= 400

Hence, the correct option is (c).

#### Page No 9.53:

#### Answer:37

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

The given A.P. is $\frac{1}{2q},\frac{1-2q}{2q},\frac{1-4q}{2q},...$

Common difference = *d* = Second term − First term

= $\frac{1-2q}{2q}-\frac{1}{2q}$

= $\frac{-2q}{2q}=-1$

Hence, the correct option is (a).

#### Page No 9.53:

#### Answer:38

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

The given A.P. is $\frac{1}{3},\frac{1-3b}{3},\frac{1-6b}{3},...$

Common difference = *d* = Second term − First term

= $\frac{1-3b}{3}-\frac{1}{3}$

= $\frac{-3b}{3}=-b$

Hence, the correct option is (c).

#### Page No 9.53:

#### Answer:39

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

The given A.P. is $\frac{1}{2b},\frac{1-6b}{2b},\frac{1-12b}{2b},...$

Common difference = *d* = Second term − First term

= $\frac{1-6b}{2b}-\frac{1}{2b}$

= $\frac{-6b}{2b}=-3$

Hence, the correct option is (d).

#### Page No 9.53:

#### Answer:27

Here, we are given the first term of the A.P. as *a* and the *n*^{th} term (*a*_{n}) 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).**

#### Page No 9.53:

#### Answer:28

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).**

#### Page No 9.53:

#### Answer:29

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 30^{th} terms.

So, let us first find the 30^{th} term of first A.P.

Similarly, we find the 30^{th} term of second A.P.

Now, the difference between the 30^{th} terms is,

Therefore,

Hence, the correct option is **(d).**

#### Page No 9.53:

#### Answer:30

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

First term (*a*_{1}) =

Second term (*a*_{2}) =

Third term (*a*_{3}) =

Fourth term (*a*_{4})=

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)**.

#### Page No 9.53:

#### Answer:31

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 *18*^{th }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 *n*^{th}* *term, we replace *n* by. We get,

As we know,

Therefore, for the 18^{th} terms, we get,

Hence

Hence no option is correct.

#### Page No 9.53:

#### Answer:32

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).**

#### Page No 9.53:

#### Answer:33

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 *n*^{th} term.

So we know that the *n*^{th}term 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).**

#### Page No 9.53:

#### Answer:34

Here, we are given an A.P. whose *n*^{th} 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 *n*^{th} term of A.P.

Now, the last term (*l*) or the *n*^{th} 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).**

#### Page No 9.53:

#### Answer:35

In the given problem, we are given an A.P whose 18^{th}^{ }and 11^{th}^{ }term are in the ratio 3:2

We need to find the ratio of its 21^{st} and 5^{th} 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 21^{st} and 5^{th} term is

Therefore the correct option is **(b)**.

#### Page No 9.6:

#### Answer:2

Here, we are given the *n*^{th }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,

#### Page No 9.6:

#### Answer:3

In the given problem, we are given the first, second term and* *the *n*^{th }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

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