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Page No 5.60:

Question 11:

Write the common difference of an A.P. whose nth term is an = 3n + 7.

Answer:

In the given problem, nth term is given by,

.

To find the common difference of the A.P., we need two consecutive terms of the A.P.

So, let us find the first and the second term of the given A.P.

First term,

Second term (),

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

Therefore, the common difference is.



Page No 224:

Question 1:

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

(i) an = 3n + 2

(ii) an = 3n

(iii)  an = (−1)n · 2n

(iv) an=n(n-2)2

(v) ann2 − n + 1

 (vi) an = 2n2 − 3n + 1

Answer:

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

(i)

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

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

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

(ii)

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

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

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

(iii)

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

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

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

(iv)

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

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

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

(v)

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

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

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

(vi)

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

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

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

Page No 224:

Question 2:

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

(i) an = 5n − 4; a12and a15

(ii) an =3n-24n + 5; a7 and a8

(iii) an = n (n −1) (− 2); a5 and a8

(iv) an = (−1)n n ; a3, a5, a8

Answer:

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

(i)

We need to find and .

Now, to find  term we use, we get,

Also, to find term we use , we get,
a15=515-4=75-4=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, .

Page No 224:

Question 3:

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

(i) a1 = 1, an = an−1 + 2, n ≥ 2

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

Answer:

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

We need to find its next five terms.

(i) 

Here, we are given that 

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

Now …… (1)

So, to find the  term 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 the  term 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

.​



Page No 226:

Question 1:

Write the sequence with nth term:

(i) an = 3 + 4n

(ii) an = 5 + 2n

(iii) an = 6 − n

(iv) an = 9 − 5n

Show that all of the above sequences form A.P.

Answer:

In the given problem, we are given the sequence with the nth 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 226:

Question 2:

The nth term of an A.P. is 6n + 2. Find the common difference.

Answer:

In the given problem, nth term is given by “”. To find the common difference of the A.P., we need two consecutive terms of the A.P.

So, let us find the first and the second term of the given A.P.

First term,

Second term (),

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

Therefore, the common difference is.

Page No 226:

Question 3:

Show that the sequence defined by an = 5n −7 is an A.P, find its common difference.

Answer:

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 227:

Question 4:

Show that the sequence defined by an = 3n2 − 5 is not an A.P.

Answer:

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 227:

Question 5:

The general term of a sequence is give by an = −4n + 15. Is the sequence an A.P.? If so, find its 15th term and the common difference.

Answer:

In the given problem, we need to find that the given sequence is an A.P or not and then find its 15thterm 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 15th using the formula

First term (a) = 11

n = 15

Common difference (d) = −4

Substituting the above values in the formula

Therefore,

Page No 227:

Question 6:

Justify whether it is true to say that the sequence having following nth term is an A.P.
(i) an = 2n − 1      (ii)  an = 3n2 + 5     (iii)  an = 1 + n + n2

Answer:

(i)

Consider the expression an = 2n − 1,
For n = 1, a1 = 2(1) − 1 = 1
For n = 2, a2 = 2(2) − 1 = 3
For n = 3, a3 = 2(3) − 1 = 5
For n = 4, a4 = 2(4) − 1 = 7
The first four terms are 1, 3, 5, 7.
The difference between each consecutive term is 2.
Hence this is an A.P.

(ii)
Consider the expression an = 3n2 + 1,
For n = 1, a1 = 3(12) + 1 = 8
For n = 2, a2 = 3(22) + 1 = 17
For n = 3, a3 = 3(32) + 1 = 32
For n = 4, a4 = 3(42) + 1 = 53
The first four terms are 8, 17, 32, 53.
The difference between each consecutive is not same.
Hence this is not an A.P.

(iii)
Consider the expression an = 1 + n + n2,
For n = 1, a1 = 1 + 1 + 1 = 3
For n = 2, a2 = 1 + 2 + 4 = 7
For n = 3, a3 = 1 + 3 + 9 = 13
For n = 4, a4 = 1 + 4 + 16 = 21
The first four terms are 3, 7, 13, 21.
The difference between each consecutive term is not same.
Hence this is not an A.P.

 

Page No 227:

Question 7:

The nth term of an A.P. cannot be n2 + 1. Justify your answer.

Answer:

Let nth term of an A.P. be n2 + 1.
an=n2+1a1=12+1=2a2=22+1=5a3=32+1=10a4=42+1=17

The common difference of an AP (d) is given by the difference of two successive terms.
a2-a1=5-2=3a3-a2=10-5=5a4-a3=17-10=7As, a2-a1a3-a2a4-a3

Thus, the nth term of an A.P. cannot be n2 + 1.



Page No 228:

Question 1:

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

(i) −5, −1, 3, 7, ...

(ii) 15,35,55,75

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

(iv) −1.1, −3.1, −5.1, −7.1, ...

Answer:

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 229:

Question 2:

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

(i) a = 4, d = −3

(ii) a = −1, d = 12

(iii) a = −1.5, d = −0.5

Answer:

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 229:

Question 3:

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

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

(ii) The amount of air present in the cylinder when a vacuum pump removes each time 14 of their remaining in the cylinder.

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

Answer:

(i) In the given problem,

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

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

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

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

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

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

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

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

Here,

Also,

Therefore,

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

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

So,

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

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

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

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

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

Here,

Also,

Since,

The sequence is not an A.P.

(iii)
Here, prinical (P) = 1000
Rate (r) = 10%
Amount compounded annually is given by
A=P1+r100n
For the first year,
A1=10001+101001=1100
For the second year,
A2=10001+101002=1210
For the third year,
A1=10001+101003=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 229:

Question 4:

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

(i) 1, −2, −5, −8, ...

(ii) 0, −3, −6, −9, ...

(iii) -1, 14, 32, ...

(iv) -1, 56, 23, ...

Answer:

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

(i)

Here, first term (a1) =1

Common difference (d)

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

That is we need to find

So, using the formula

Substituting in the above formula

Substituting n = 5, we get

Substituting n = 6, we get

Substituting n = 7, we get

Substituting n = 8, we get

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

(ii)

Here, first term (a1) =0

Common difference (d)

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

That is we need to find

So, using the formula

Substituting in the above formula

Substituting n = 5, we get

Substituting n = 6, we get

Substituting n = 7, we get

Substituting n = 8, we get

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

(iii)

Here, first term (a1) =−1

Common difference (d)

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

That is we need to find

So, using the formula

Substituting in the above formula

Substituting n = 4, we get

Substituting n = 5, we get

Substituting n = 6, we get

Substituting n = 7, we get

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

(iv)

Here, first term (a1) =−1

Common difference (d)

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

That is we need to find

So, using the formula

Substituting in the above formula

Substituting n = 4, we get

a4=-1+4-116a4=-1+12a4=-2+12=-12

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 229:

Question 5:

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

(i) 3, 3, 3, 3, ...

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

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

(iv) −225, −425, −625, −825, ...

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

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

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

Answer:

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 an A.P and its common difference is 

(ii) 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 .

(iii) 

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 

(iv) 

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 .

(v) 

Here,

First term (a) = 10

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

Common difference (d) = a1-a=a2-a1 

Here,


Also,


Since 

Hence, the given sequence is not an A.P

(vi) 

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.

(vii) 

Here,

First term (a) = + b

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

Common difference (d

Here,

Also,

Since 

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

Page No 229:

Question 6:

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

Answer:

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

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

Here,

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

So,

Substituting n = 1, we get

Substituting n = 2, we get

Substituting n = 3, we get

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

Common difference (d)

Here,

Also,

Since

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



Page No 238:

Question 1:

Find:

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

(ii) 18th term of the A.P. 2, 32, 52...

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

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

Answer:

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

(i) 10th term of the A.P. 

Here,

First term (a) = 1

Common difference of the A.P. (d
                                                         

Now, as we know,

So, for 10th term,

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

(ii) 18th term of the A.P. 

Here,

First term (a) = 

Common difference of the A.P. (d
                                                         

Now, as we know,

So, for 18th term,

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

(iii) nth term of the A.P. 

Here,

First term (a) = 13

Common difference of the A.P. (d
                                                         

Now, as we know,

So, for nth term,

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

(iv) 10th term of the A.P. 

Here,

First term (a) = −40

Common difference of the A.P. (d
                                                        

Now, as we know,

So, for 10th term,

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

Page No 238:

Question 2:

​Find:

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

(ii) Which term of the A.P. 84, 80, 76, ... is 0?

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

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

Answer:

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 

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.

(iv)  –7, –12, –17, –22,... 

Given:a=-7d=-12--7  =-12+7  =-5an=-82Now,an=a+n-1d-82=-7+n-1-5-82=-7-5n+5-82=-2-5n-82+2=-5n-80=-5nn=-80-5n=16Hence, 16th term of the A.P. is -82.If an=-100an=a+n-1d-100=-7+n-1-5-100=-7-5n+5-100=-2-5n-100+2=-5n-98=-5nn=-98-5n=19.6 which is not possible since n is a natural numberHence, -100 is not the term of this A.P.​

 

Page No 238:

Question 3:

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

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

Answer:

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, 302 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

Since, the value of n is a fraction.

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

Therefore, the answer is .

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Question 4:

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

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

(ii) -1,-56,-23,-12,...103.

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

(iv) 18, 1512, 13, ..., -47.

Answer:

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 238:

Question 5:

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

Answer:

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 238:

Question 6:

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

Answer:

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

We need to find the 40th term

Here,

Now, we will find and using the formula

So,

Also,

So, to solve for a and d

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

Substituting (3) in (1), we get

Thus,

Substituting the above values in the formula

Therefore,

Page No 238:

Question 7:

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

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

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

Answer:

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

(i) 3, 8, 13 …253

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

So,

First term (a) = 3

Last term (an) = 253

Common difference, d = 
                                     

Now, as we know,

So, for the last term,

Further simplifying,

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

So, for the 40th term (n = 40)

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

(iii) 1, 4, 7, 10 …88

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

So,

First term (a) = 1

Last term (an) = 88

Common difference, 
                                     = 3

Now, as we know,

So, for the last term,

Further simplifying,

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

So, for the 19th term (n = 19)

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

Page No 238:

Question 8:

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

Answer:

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

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

According to the question,

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

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

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

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

Page No 238:

Question 9:

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

Answer:

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

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

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

According to the question,

⇒ 0 = 40 + (n − 1)(−3)
⇒ 0 = 40 − 3n + 3
⇒ 3n = 43
⇒ n = 433          .... (1)

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

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

Page No 238:

Question 10:

If the seventh term of an A.P. is 19 and its ninth term is 17, find its (63)rd term.

Answer:

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

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

According to the question,

a7 = 19
⇒ a + (7 − 1)d = 19
⇒ a + 6d = 19       .... (1)

Also, a9 = 17
⇒ a + (9 − 1)d = 17
⇒ a + 8d = 17   ....(2)

On Subtracting (1) from (2), we get
8− 6d = 17-19
⇒ 29-763
⇒ 2= 263
⇒ 163
⇒ a = 19-663     [From (1)]
⇒ a = 7-663
⇒ a = 163

∴ a63 a + (63 − 1)d
        =
 163+6263
        = 6363
        = 1

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

Page No 238:

Question 11:

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

Answer:

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

A.P. is 3, 15, 27, 39 …

Here,

First term (a) = 3

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

Now, as we know,

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

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

Also,

Further simplifying, we get,

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



Page No 239:

Question 12:

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

Answer:

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

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

First A.P. is 63, 65, 67 …

Here,

First term (a) = 63

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

Now, as we know,

So, for nth term,

Second A.P. is 3, 10, 17 …

Here,

First term (a) = 3

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

Now, as we know,

So, for nth term,

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

Therefore,

Page No 239:

Question 13:

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

Answer:

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

Now, as we know,

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

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

For 7th term (n = 7),

Now, we are given,

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

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

Therefore, the A.P. is

Page No 239:

Question 14:

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

Answer:

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

We can write this as,

We need to find a and d

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

As we know,

For 4th term (n = 4),

For 8th term (n = 8),

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

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

For 10th term (n = 10),

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

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

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

On further simplifying, we get,

Therefore, for the given A.P

Page No 239:

Question 15:

The sum of the 5th and 7th terms of an A.P. is 52 and the 10th term is 46. Find the A.P

Answer:

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

The nth term of the AP is given as:
an = a + (n − 1)d
a5+a7=52a+4d+a+6d=522a+10d=52a+5d=26  .....ia10=46a+9d=46 .....ii

From (i) and (ii), we get
a = 1, d = 5

Thus, the A.P. is 1, 6, 11, 16, ... .

Page No 239:

Question 16:

​​If the sum of 3rd and 8th terms of an A.P is 7 and the sum of the 7th and the 14th terms is –3, find the 10th term. 

Answer:

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

nth term, an = a + (n − 1)d
a3+a8=7a+2d+a+7d=72a+9d=7 .....iAnd, a7+a14=-3a+6d+a+13d=-32a+19d=-3 .....ii

From (i) and (ii), we get
a = 8, d = −1

a10=a+9d=8+9-1=-1

Hence, the 10th term of the A.P. is −1.

Page No 239:

Question 17:

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

Answer:

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

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

So, let us first find the two terms.

So, as we know,

For 10th term (n = 10),

For 15th term (n = 15),

Now, we are given,

Solving this, we get,

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

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

Hence proved

Page No 239:

Question 18:

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

Answer:

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

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

So, as we know,

We get,

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

For 19th term (n = 19),

(Using 1)

For 29th term (n = 29),

a29=a+29-1d=-8d+28d=20d=2×10d=2×a19  (Using 1)

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

Hence proved.

Page No 239:

Question 19:

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

Answer:

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

We have to prove that

So, let us first find the two terms.

As we know,

For 10th term (n = 10),

For 24th term (n = 24),

Now, we are given that

So, we get,

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

For 34th term (n = 34),

(Using 1)

For 72nd term (n = 72),

(Using 1)

Therefore,

Page No 239:

Question 20:

The 26th , 11th  and last term of an A.P. are 0, 3 and -15, respectively. Find the common difference and the  number of terms .

Answer:

It is given that, a26=0, a11=3, an=-15.
a+25d=0       .....(i)a+10d=3     .....(ii)Subtracting (ii) from (i).15d=-3d=-15a=-25d=5Now, an=a+n-1d-15=5+n-1-15-15-5=n-1-15n=27

Page No 239:

Question 21:

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

Answer:

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

Here, we are given that,

We need to find a and d

 

So, as we know,

For the 4th term (n = 4),

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

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

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

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

On further simplification, we get,

Now, to find a,

Therefore, for the given A.P

Page No 239:

Question 22:

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

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

(ii) a10 −a5= 200

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

Answer:

A.P: a, a+d, a+2d …

Here, we first need to write the expression for

Now, as we know,

So, for nth term,

Similarly, for kth term

So,

So,

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

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

Here,

Now, we will find and using the formula

So,

Also,

Solving for a and d

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

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

(ii) We are given,

Here,

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

Now, as we know,

Here, we find a30 and a20.

So, for 10th term,

Also, for 5th term,

So,

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

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

Here

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

Now, as we know,

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

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

Now, we are given,

On substituting the values, we get,

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

Page No 239:

Question 23:

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

Answer:

a8=12a2 and a11=13a4+1a+7d=a+d 2a+13d=0  .....(i)
And,  a11=13a4+1 a+10d=a+3d3+12a+27d=1 .....(ii)Solve (i) and (ii), we get d = 1.Therefore, a = -13a15=a+14d=-13+14=1

Page No 239:

Question 24:

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

Answer:

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

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

We need to find the difference between their 100th terms

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

Now, as we know,

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

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

Now, we are given,

On substituting the values, we get,

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

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

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

So,

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

Page No 239:

Question 25:

How many multiples of 4 lie between 10 and 250?

Answer:

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

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

So here,

First term (a) = 12

Last term (an) = 248

Common difference (d) = 4

So, let us take the number of terms as n

Now, as we know,

So, for the last term,

Further simplifying,

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

Page No 239:

Question 26:

How many three digit numbers are divisible by 7?

Answer:

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

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

So here,

First term (a) = 105

Last term (an) = 994

Common difference (d) = 7

So, let us take the number of terms as n

Now, as we know,

So, for the last term,

Further simplifying,

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

Page No 239:

Question 27:

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

Answer:

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

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

According to the question,

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

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

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

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

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

Page No 239:

Question 28:

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

Answer:

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

Next number is 108 + 9 = 117.

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

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

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

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

According to the question,

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

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

Page No 239:

Question 29:

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

Answer:

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

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

According to the question,

a19 = 3a6
⇒ a + (19 − 1)d =  3(a + (6 − 1)d)
⇒ a + 18d =  3a + 15d
⇒ 18d − 15=  3a − a
⇒ 3= 2a
⇒ a = 32d   .... (1)

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

On substituting the values of (1) in (2), we get
32+ 8d = 19
⇒ 3+ 16= 19 × 2
⇒ 19= 38
⇒ = 2
⇒ a = 32×2     [From (1)]
⇒ a = 3

Thus, the A.P. is 3, 5, 7, 9, .... .

Page No 239:

Question 30:

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

Answer:

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

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

Next number is 110 + 10 = 120.

The last number that is divisible by 10 is 990.

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

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

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

According to the question,

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

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

Page No 239:

Question 31:

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

Answer:

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

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

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

According to the question,

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

Therefore, total number of terms is 23.

Since, there are odd number of terms.
So, Middle term will be 23+12th term, i.e., the 12th term.

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

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

Page No 239:

Question 32:

 Find the sum of two middle terms of the A.P. : -43, -1, -23, -13,......., 413.

Answer:

a=-43, d=13, an=133a+n-1d=133-43+n-113=13313=-4+n-1n=18Midlle terms are n2th and  n2+1th , i.e 9th and 10th terms.a9=a+8d=-43+83=43a10=a+9d=-43+93=53a9+a10=93=3

Page No 239:

Question 33:

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

Answer:

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

We need to prove that

So, let us first find the two terms.

As we know,

For (m+1)th term (n’ = m+1)

For (n+1)th term (n’ = n+1),

Now, we are given that

So, we get,

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

For (m+n+1)th term (n’ = m+n+1 ),

(Using 1)

For (3m+1)th term (n’ = 3m+1),

Therefore,

Hence proved

Page No 239:

Question 34:

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

Answer:

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

Here,

The first term (a) = a

The last term (an) = l

Now, as we know,

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

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

So, we get,

Now, we need to prove

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

Therefore,

Hence proved

Page No 239:

Question 35:

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

Answer:

The given A.P is 11, 15, 19, .....,299.
a=11d=4an=299a+n-1d=29911+n-14=299n-14=288n=73



Page No 240:

Question 36:

For the A.P. : -3, -7, -11 ,....., can we find a30  - a20 withoput actually finding a30 and  a20  ? Give reasons for your answer.

Answer:

Consider the A.P −3, −7, −11, ....
a=-3, d=-4, a30-a20=a+30-1d-a+20-1da30-a20=a+29d-a-19da30-a20=10d=10-4=-40

Page No 240:

Question 37:

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

Answer:

First term of 1st A.P is 2.
First term of 2nd A.P is 7.
Consider the difference of their 10th terms,
a10-a'10=a+9d-a'-9d'=a-a'+9d-9d'=2-7+0               d=d'=-5a21-a'21=a+20d-a'-20d'=a-a'+20d-20d'=2-7+0               d=d'=-5Therefore, a10-a'10=a21-a'21.
The difference between any two corresponding terms of A.P's is same as the difference between their terms.



Page No 242:

Question 1:

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

Answer:

Here, we are given three terms,

First term (a1) =

Second term (a2) =

Third term (a3) =

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

…… (1)

Also,

…… (2)

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

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

Page No 242:

Question 2:

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

Answer:

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

First term (a1) =

Second term (a2) =

Third term (a3) =

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

…… (1)

Also,

…… (2)

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

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

Page No 242:

Question 3:

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

Answer:

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

First term (a1) =

Second term (a2) =

Third term (a3) =

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

…… (1)

Also,

…… (2)

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

Hence proved

Page No 242:

Question 4:

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

Answer:

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 242:

Question 5:

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

Answer:

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 242:

Question 6:

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

Answer:

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 242:

Question 7:

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

Answer:

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 242:

Question 8:

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

Answer:

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

Also, a-3da+3da-da+d=56a2-9d2a2-d2=56142-9d2142-d2=56196-9d2196-d2=56

1176-54d2=980-5d2196=49d2d2=4d=±2

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



Page No 243:

Question 9:

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

Answer:

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

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

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

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

a-da+d=5da2-d2=5d62-d2=5d36-d2=5dd2+5d-36=0d2+9d-4d-36=0dd+9-4d+9=0d+9d-4=0d=4, -9For d=4,The numbers are 6-4, 6, 6+4i.e. 2, 6, 10For d=-9,The numbers are 6+9, 6, 6-9i.e. 15, 6, -3

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

Page No 243:

Question 10:

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

Answer:

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 243:

Question 11:

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

Answer:

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

a-d+a+a+d=2073a=207a=69Now, a-d×a=46236969-d=462369-d=67d=2Therefore, the three required parts are 67, 69 and 71.

 

Page No 243:

Question 12:

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

Answer:

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

a-d+a+a+d=180  angle sum property3a=180a=60Now, a+d=2a-da+d=2a-2da=3dd=603=20Therefore, the three angles of a triangle are 40, 60, 80.



Page No 258:

Question 1:

Find the sum of the following arithmetic progressions:
​
(i) −26, −24, −22, ... to 36 terms.

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

(iii) (xy)2, (x2 + y2), (x + y)2, ..., to n terms

(iv) x-yx+y, 3x-2yx+y, 5x-3yx+y, ... to n terms
 

Answer:

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.

= number of terms

(i)  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.

(ii)  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.

(iii) 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 .

(iv)  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.

Page No 258:

Question 2:

Which term of the A.P.  -2, -7, -12,.... will be -77 ? Find the sum of this A.P. upto the term -77 .

Answer:

Here, a = −2 and d = −5.
If an=-77.-2+n-1-5=-77-2-5n+5=-77n=16Therefore,Sn=n22a+n-1dS16=1622-2+16-1-5=8-4-75=-632

Page No 258:

Question 3:

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

Answer:


The given A.P is 8, 10, 12, 14,...., 126.
a = 8 and d = 2.
When this A.P is reversed, we get the A.P.
126, 124, 122, 120,....
So, first term becomes 126 and common difference −2.
The sum of first 10 terms of this A.P is as follows:
S10=1022×126+9-2=5234=1170

Page No 258:

Question 4:

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

(i) a= 3 + 4n

(iv) yn = 9 − 5n

Answer:

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

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

Where, a = the first term

l = the last term

So, for the given A.P,

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

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

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

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

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

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

Where, a = the first term

l = the last term

So, for the given A.P,

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

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

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

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

Page No 258:

Question 5:

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

Answer:

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

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

Where, a = the first term

l = the last term

So, for the given A.P,

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

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

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

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

Page No 258:

Question 6:

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

Answer:

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,

3n2 - 53n + 232 = 0

On solving by splitting the middle term, we get,

3n2 -24n -29n +232 = 03nn - 8 - 29n - 8 = 03n - 29n - 8 = 0

Further,

Also,

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

So, the term is a8

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



Page No 259:

Question 7:

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

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

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

Answer:

​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 636. For that, we will use the formula,

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

d = common difference of the given A.P.

= number of terms

The first term (a) = 9

The sum of n terms (Sn) = 636

Common difference of the A.P. (d) = 

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

So, we get the following quadratic equation,

On solving by splitting the middle term, we get,

Further,

Or,

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

(ii) 27, 24, 21... 

Given:a=27d=24-27  =-3Sn=0Now,Sn=n22a+n-1d0=n2227+n-1-30=n227+n-1-3n54-3n+3=0n57-3n=0n=0 or 3n=57But n is a natural number,  n0n=19

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

(iii) The AP is as follows: 45, 39, 33...
a=45d=39-45=-6Sn=180Now,Sn=n22a+n-1d180=n2245+n-1-6180=n45-3n+3n2-16n+60=0n-6n-10=0n=6 or n=10

Now, the AP obtained is as follows: 45, 39, 33, 27, 21, 15, 9, 3, −3, −9,...
The 6th term of the AP is 15 while the 10th one is −9. We observe that the terms after the 6th term till the 10th one cancel out each other and thus, the sum of first 6 terms of the AP is equal to that of the first 10 terms.

Page No 259:

Question 8:

Find the sum of

(i) the first 15 multiples of 8

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

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

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

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

(vi) first 8 multiples of 3.

Answer:

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)11990=110+11n-11990=99+11n891=11n81=n

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

Sn=8122(110)+81-111Sn=812220+80×11Sn=812×1100Sn=81×550Sn=44550

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

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


96=12+(n-1)484=(n-1)421=n-122=n
Now,

First term (a) = 12

Number of terms (n) = 22

Common difference (d) =4

Now, using the formula for the sum of n terms, we get
S22=2222(12)+(22-1)4S22=1124+84S22=1188

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

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

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

Sn=n2a+lSn=823+24S8 = 4×3+24 = 108

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

Page No 259:

Question 9:

Find the sum:

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

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

(iii) 18 + 1512+13+...+ -4912​

Answer:

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.

= number of terms

(i) 

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.

(ii) 

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.
(iii) 18+1512+13+...+-4912

Common difference of the A.P. (d) = 
                                                       =1512-18=312-18=31-362=-52

So here,

First term (a) = 18

Last term (l) = -4912=-992

Common difference (d) = -52

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,

-992=18+n-1-52-992=18+-52n+5252n=18+52+99252n=18+1042n=28

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

Sn=2822×18+28-1-52Sn=1436+27-52Sn=-441

Therefore, the sum of the A.P is Sn=-441.

Page No 259:

Question 10:

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

Answer:

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 259:

Question 11:

Find the sum of n terms of the series 4-1n+4-2n+4-3n+..........

Answer:

Let the given series be S = 4-1n+4-2n+4-3n+..........
=4+4+4+...-1n+2n+3n+...=41+1+1+...-1n1+2+3+...=S1-S2
S1=41+1+1+...a=1, d=0S1=4×n22×1+n-1×0                  Sn=n22a+n-1dS1=4n
S2=1n1+2+3+...a=1, d=2-1=1S2=1n×n22×1+n-1×1=122+n-1=121+n
Thus, S=S1-S2=4n-121+nS=8n-1-n2=7n-12
Hence, the sum of n terms of the series is 7n-12.

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Question 12:

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

Answer:

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

Here, we are given that,

Also, we know,

For the 2nd term (n = 2),

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

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

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

So, for the given A.P

So, to find the sum of first 51 terms of this A.P., we use the following formula for the sum of n terms of an A.P.,

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

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 51, we get,

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

Page No 259:

Question 13:

The sum of first n terms of an A.P is 5n2 + 3n. If its mth terms is 168, find the value of m. Also, find the 20th term of this A.P.

Answer:

Sn=5n2+3nWe knowan=Sn-Sn-1an=5n2+3n-5n-12-3n-1an=10n-2

Now,

am=16810m-2=16810m=170m=17

a20=1020-2=198

Page No 259:

Question 14:

If the sum of first n terms of an A.P. is 12(3n2 + 7n), then find its nth term. Hence write its 20th term.

Answer:

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

We know that, sum of first n terms = Sn n2[2a + (n − 1)d]

It is given that sum of the first n terms of an A.P. is 12(3n2 + 7n).

∴ First term = =  S12[3(1)2 + 7(1)] = 5.

Sum of first two terms = S12[3(2)2 + 7(2)] = 13.

∴ Second term = S2 − S1 = 13 − 5 = 8.

∴ Common difference = d = Second term − First term
                                           = 8 − 5 = 3
 
Also, nth term = an = a + (n − 1)d
⇒ an = 5 + (n − 1)(3)
⇒ an = 5 + 3n − 3
⇒ an = 3+ 2

Thus, nth term of this A.P. is 3+ 2.

Now, 
a20 = a + (20 − 1)d
⇒ a20 = 5 + 19(3)
⇒ a20 = 5 + 57
⇒ a20 = 62

Thus, 20th term of this A.P is 62.

Page No 259:

Question 15:

If Sn denotes the sum of first n terms of an A.P., prove that S12 = 3(S8 − S4).

Answer:

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

We know that, sum of first n terms = Sn n2[2a + (n − 1)d]

Now,
S42[2a + (4 − 1)d]
     = 2(2a + 3d)
     = 4a + 6d              ....(1)


S82[2a + (8 − 1)d]
     = 4(2a + 7d)
     = 8a + 28d            ....(2) 


S12 122[2a + (12 − 1)d]
      = 6(2a + 11d)
      = 12a + 66d          ....(3)

On subtracting (1) from (2), we get
S8 − S8a + 28d − (4a + 6d)
           4a + 22d

Multiplying both sides by 3, we get

3(S8 − S4) = 3(4a + 22d)
                 = 12a + 66d
                 = S
12                 [From (3)]

Thus, S12 = 3(S8 − S4).

Page No 259:

Question 16:

The sums of first n terms of three A.P. are S1,S2 and S3. The first term of each is 5 and their common differences are 2, 4 and 6 respectively. Prove that S1 + S3 = 2S2

Answer:

Sn=n22a+n-1dFor S1, when a=1 and d=1.S1=n221+n-1=n22+n-1=n2n+1For S2, when a=1 and d=2.S2=n221+n-12=n22+2n-2=n22n=n2For S3, when a=1 and d=3.S3=n221+n-13=n22+3n-3=n23n-1Therefore, S1+S3=n2n+1+n23n-1=n24n=2n2S1+S3=2S2
Hence proved.

Page No 259:

Question 17:

If Sn denotes the sum of the first n terms of an A.P., prove that S30 = 3(S20 − S10)

Answer:

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

We know that, sum of first n terms = Sn n2[2a + (n − 1)d]

Now,
S10 102[2a + (10 − 1)d]
      = 5(2a + 9d)
      = 10a + 45d          ....(1) 

S20 202[2a + (20 − 1)d]
      = 10(2a + 19d)
      = 20a + 190d        ....(2) 

S30 302[2a + (30 − 1)d]
      = 15(2a + 29d)
      = 30a + 435d        ....(3)

On subtracting (1) from (2), we get
S20 − S10 = 20a + 190d − (10a + 45d)
                = 10a + 145d

On multiplying both sides by 3, we get
3(S20 − S10) = 3(10a + 145d)
                    = 30a + 435d
                    = S
30                   [From (3)]

Hence, S30 = 3(S20 − S10)

Page No 259:

Question 18:

A man saved Rs 16500 in ten years. In each year after the first he saved Rs 100 more than he did in the preceding year. How much did he save in the first year?

Answer:

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 259:

Question 19:

A man saved Rs. 32 during the first year, Rs 36 in the second year and in this way he increases his saving by Rs 4 every year. Find in what time his saving will be Rs. 200.

Answer:

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 259:

Question 20:

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

Answer:

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

Here, we are given that,

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

Also, we know,

For the 3th term (n = 3),

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

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

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

So, for the given A.P

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

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

d = common difference of the given A.P.

n = number of terms

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

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

Page No 259:

Question 21:

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

Answer:

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

Here, we are given that,

Also, we know,

For the 12th term (n = 12),

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

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

d = common difference of the given A.P.

n = number of terms

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

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

On further simplifying for d, we get,

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

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

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

Page No 259:

Question 22:

(i) If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of n terms.

(ii) If the sum of first four terms of an A.P. is 40 and that of first 14 terms is 280. Find the sum of its first n terms.

Answer:

(i) 

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

Here, we are given that,

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

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

d = common difference of the given A.P.

= number of terms

So, using the formula for n = 7, we get,

Further simplifying for a, we get,

Also, using the formula for n = 17, we get,

Further simplifying for a, we get,

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

Now, to find a, we substitute the value of d in (3),

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

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

(ii) Given:
S4=40S14=280Now,Sn=n22a+n-1dS4=422a+4-1d40=22a+3d2a+3d=20            ...(1)S14=1422a+14-1d280=72a+13d2a+13d=40            ...(2)Subtracting (1) from (2), we get10d=20d=2Substituting the value of d in (1), we geta=7Therefore, a=7 and d=2Thus,Sn=n22×7+n-12     =n214+2n-2     =n212+2n     =n6+n     =n2+6n

Hence,  the sum of its first n terms is n2 + 6n.

Page No 259:

Question 23:

The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Answer:

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 260:

Question 24:

​The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find its common difference.

Answer:

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

We know that, sum of first n terms = Sn n2[2a + (n − 1)d]

Also, nth term = an = a + (n − 1)d

According to the question,

a = 7, an = 49 and Sn = 420

Now,
an = a + (n − 1)d
 49 = 7 + (n − 1)d
​⇒ 42 = nd − d
​⇒
 nd − = 42                     ....(1)

Also,
Sn n2[2 × 7 + (n − 1)d]
⇒ 420 = n2[14 + nd − d]
⇒ 840 = n[14 + 42]               [From (1)]
⇒ 56n = 840
⇒ n = 15                              ....(2) 

On substituting (2) in (1), we get

nd − = 42
⇒ (15 − 1)= 42
⇒ 14= 42
⇒ = 3

Thus, common difference of the given A.P. is 3.

Page No 260:

Question 25:

The sum of first 9 terms of an A.P. is 162. The ratio of its 6th term to its 13th term is 1 : 2. Find the first and 15th term of the A.P.


Disclaimer: There is a misprint in the question, 'the sum of 9 terms' should be witten instead of 'the sum of q terms'.

Answer:

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

We know that, sum of first n terms = Sn n2[2a + (n − 1)d]

Also, nth term = an = a + (n − 1)d

According to the question,

Sq = 162 and a6a13=12

Now,
a6a13=12a+(6-1)da+(13-1)d=12a+5da+12d=122a+10d=a+12d2a-a=12d-10da=2d                  .....(1)

Also,
S92[2a + (9 − 1)d]
⇒ 162 = 92[2(2d) + 8d]           [From (1)]
⇒ 18 = 12[12d]              
⇒ 18 = 6d
⇒ d = 3
⇒ a = 2 × 3                              [From (1)]
⇒ a = 6

Thus, the first term of the A.P. is 6.

Now,

a15 = 6 + (15 − 1)3
     = 6 + 42
    = 48

Thus, 15th term of the A.P. is 48.

Page No 260:

Question 26:

The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P.

Answer:

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

We know that, sum of first n terms = Sn n2[2a + (n − 1)d]

It is given that sum of the first 7 terms of an A.P. is 63.
And sum of next 7 terms is 161.
∴ Sum of first 14 terms = Sum of first 7 terms + sum of next 7 terms
                                     = 63 + 161 = 224

Now,
S72[2a + (7 − 1)d]
⇒ 63 = 72(2a + 6d)
⇒ 18 = 2a + 6d
⇒ 2a + 6d = 18         ....(1) 

Also,
S14 142[2a + (14 − 1)d]
⇒ 224 = 7(2a + 13d)
⇒ 32 = 2a + 13d
⇒ 2a + 13d =  32      ....(2) 

On subtracting (1) from (2), we get
13d − 6d = 32 − 18
⇒ 7d = 14
d = 2
⇒ 2a = 18 − 6d         [From (1)]
⇒ 2a = 18 − 6 × 2
⇒ 2a = 18 − 12
⇒ 2a = 6
a = 3

Also, nth term = an = a + (n − 1)d
⇒ a28 = 3 + (28 − 1)2
           = 3 + 27 × 2
           = 57

Thus, 28th term of this A.P. is 57.

Page No 260:

Question 27:

The nth term of an A.P is given by (−4n + 15), Find the sum of first 20 terms of this A.P.

Answer:

an=-4n+15a1=-4+15=11Also, a2=-8+15=7Common difference, d=a2-a1=7-11=-4Now,S20=2022×11+20-1-4=1022-76=-540

Page No 260:

Question 28:

Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25th term.

Answer:

First term, a = 10

Sum of first 14 terms, S14=1505

1422×10+14-1d=15057×20-13d=150520-13d=15057=21513d=-195d=-15

Now,
a25=10+24-15=-350

Page No 260:

Question 29:

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

​(ii) The first and last terms of an A.P. are –5 and 45 respectively. If the sum of the terms of the A.P. is 120, then find the number of terms and the common difference.

Answer:

(i) 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 .

(ii)

The first term of the A.P (a) = –5

The last term of the A.P (l) = 45

The sum of the terms of the A.P. = 120

Let the number of terms be n.

So, as we know that,

S=n2a+l120=n2-5+45120=n2×40n=6
 

Also,

45=-5+6-1d5d=50d=505d=10

Therefore, the number of terms is 6 and the common difference is 10.

Page No 260:

Question 30:

If the sum of the first n terms of an A.P is 4n − n2, What is the first term? What is the sum of first two terms? What is the second term? Similarly, find the third, the tenth and the nth terms.

Answer:

In the given problem, the sum of n terms of an A.P. is given by the expression,

So here, we can find the first term by substituting,

Similarly, the sum of first two terms can be given by,

Now, as we know,

So,

Now, using the same method we have to find the third, tenth and nth term of the A.P.

So, for the third term,

Also, for the tenth term,

So, for the nth term,

Therefore, .

Page No 260:

Question 31:

(i) If the sum of first n terms of an A.P. is n2, then find its 10th term.
(ii) If an = 3 – 4n, show that a1, a2, a3,... form an A.P. Also, find S20.
​(iii) In an A.P., if Sn = n(4n + 1), find the A.P.

Answer:

​(i) Given:
Sn=n2Sum of one term=S1=12a1=a=1      ...1Sum of two terms=S2=22a1+a2=4       ...2Subtracting 1 from 2, we geta2=3a+d=3d=2       a=1Therefore, a10=a+10-1d      =1+92      =19

Hence, the 10th term is 19.

(ii) an = 3 – 4n
a1=3-4×1=-1a2=3-4×2=-5a3=3-4×3=-9a4=3-4×4=-13a2-a1=-5--1=-4a3-a2=-9--5=-4a4-a3=-13--9=-4

Since, the difference of consecutive terms is same, it is an A.P.

Sn=n2a+an=n2-1+3-4n=n22-4n=n1-2nS20=20×1-40=-20×39=-780

Hence, S20 is −780.

(iii) Sn = n(4n + 1)

Sn-1=n-14n-1+1=n-14n-3

an=Sn-Sn-1=n4n+1-n-14n-3=4n2+n-4n2+3n+4n-3=8n-3

a1=8×1-3=5a2=8×2-3=13a3=8×3-3=21a4=8×4-3=29

Hence, the A.P. is 5, 13, 21, 29, ....

Page No 260:

Question 32:

Find the sum of first seven numbers which are multiples of 2 as well as of 9.

Answer:

The first seven numbers which are multiples of 2 as well as 9 form the AP i.e, 18, 36, 54, .....

First term = a = 18,
common difference = d = 36 -18 = 18,
number of terms = n = 7

The sum of the 7 terms is:
Sn=n22a+n-1dS7=722×18+7-1×18=7236+108=72×144=504

Hence, the sum of the first seven numbers, which are multiples of 2 and 9 is 504.

Page No 260:

Question 33:

Find the sum of all odd numbers between (i) 0 and 50 (ii) 100 and 200.

Answer:

(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 260:

Question 34:

Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.

Answer:

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 260:

Question 35:

If the sum of first 6 terms of an A.P. is 36 and that of first 16 terms is 256, find the sum of first 10 terms.

Answer:

The sum of first terms of an AP is Sn=n22a+n-1d.

Given that,
 S6=36622a+6-1d=3632a+5d=36  6a+15d=36                 .....1
and S16 = 256
1622a+16-1d=25682a+15d=25616a+120d=256               .....2

Solving equations (1) and (2),
16a-48a=256-288-32a=-32a=1

From (1),
15d=36-6=30
d=2

Now, the sum of 10 terms is given as: 
S10=1022×1+10-1×2=52+18=100
Hence, the sum of 10 terms of the AP is 100.

Page No 260:

Question 36:

​Find the sum of first 17 terms of an A.P. whose 4th and 9th terms are –15 and –30 respectively.

Answer:

Given that, a4=-15 and a9=-30.

The nth term of an AP is given by an=a+n-1d.

a4=-15a+3d=-15                  .....1and a9=-30a+8d=-30                 .....2

Subtracting (1) from (2),
a+8d-a+3d=-30--15a+8d-a-3d=-155d=-15d=-3

From (1),
a=-15-3-3=-15+9=-6

The sum of the first 17 terms of the AP
S17=1722×-6+17-1×-3                      Sn=n22a+n-1d=172-12-48=172×-60=-510
Hence, the sum of first 17 terms of an AP is -510.

Page No 260:

Question 37:

Find the sum of all integers between 50 and 500, which are divisible by 7.

Answer:

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 260:

Question 38:

A thief, after commiting a theft runs at a uniform speed of 50 m/minute . After 2 minutes, a policeman runs to catch him. He goes 60 m in first minute and increases his speed by 5 m/minute everey suceeding minute. After how many minutes, the policeman will catch the thief.?

Answer:

Suppose the policeman catches the thief after t minutes.

Uniform speed of the thief = 50 m/min

∴ Distance covered by thief in (t + 2) minutes = 50 m/min × (t + 2) min = 50 (t + 2) m

The distance covered by the policeman in t minutes is in AP, with 60 and 5 as the first term and the common difference, respectively.

Now,

Distance covered by policeman in t minutes = Sum of t terms

=t22×60+t-1×5=t2115+5t m

When the policeman catches the thief, we have

t2115+5t=50t+2115t+5t2=100t+2005t2+15t-200=0t2+3t-40=0t+8t-5=0

So, t = −8 or t = 5

∴ t = 5  (As t cannot be negative)

Thus, the policeman catches the thief after 5 min.

Page No 260:

Question 39:

Resham wanted to save at least â‚¹6500 for sending her daughter to school next year (after 12 months). She saved â‚¹450 in the first month and raised her savings by â€‹â‚¹20 every next month. How much will she able to save in next 12 months ? Will she be able to send her daughter to the school next year ?

Answer:

It is given that Reshma saved ₹450 in the first month and raised her savings by ₹20 every next month.
So, her savings are in an AP, with the first term (a) = ₹450 and the common difference (d) = ₹20.
We need to find her savings for 12 months, so n = 12.
We know that the sum of n terms of an AP is Sn=n22a+n-1d.
Reshma's savings for 12 months:
S12=1222×450+12-1×20       =6900+220       =6×1120       =6720
So, she will save ₹6,720 in 12 months.
She needed to save at least ₹6,500 for sending her daughter to school next year.
Since ₹6,720 is greater than â‚¹6,500, Reshma can send her daughter to school.
The question aims to encourage personal savings and emphasise the need of female education.

Page No 260:

Question 40:

Ramkali would need â‚¹1800 for admission fee and books etc., for her daughter to start going to school from next year. She saved â‚¹50 in the first month of this year and increased her monthly saving by â‚¹20. After a year, how much money will she save? Will she be able to fulfil her dream of sending her daughter to school?

Answer:

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

We know that, sum of first n terms = Sn n2[2a + (n − 1)d]

According to the question,
Saving of Ramkali in 1 year = â‚¹50 + â‚¹70 + â‚¹90.......

Here, a = 50, d = 70 − 50 = 20 and n = 12.

∴ S12 = 122[2 × 50 + (12 − 1)20]
          = 6[100 + 220]
          = 6 × 320
         = 1920

Hence, After a year, she will save â‚¹1920.

Since, required amount for admission is â‚¹1800 and her savings will be â‚¹1920.

Thus, yes she will be able to fulfil her dream of sending her daughter to school.

Page No 260:

Question 41:

A piece of equipment cost a certain factory Rs 60,000. If it depreciates in value, 15% the first, 13.5% the next year, 12% the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost?

Answer:

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 260:

Question 42:

A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each prize.

Answer:

In the given problem,

Total amount of money (Sn) = Rs 700

There are a total of 7 prizes and each prize is Rs 20 less than the previous prize. So let us take the first prize as Rs a.

So, the second prize will be Rs, third prize will be Rs.

Therefore, the prize money will form an A.P. with first term a and common difference −20.

So, using the formula for the sum of n terms,

We get,

On further simplification, we get,

Therefore, the value of first prize is Rs 160.

Second prize = Rs 140

Third prize = Rs 120

Fourth prize = Rs 100

Fifth prize = Rs 80

Sixth prize = Rs 60

Seventh prize= Rs 40

So the values of prizes are



Page No 261:

Question 43:

Solve the question -4 + -1 + 2 + 5 +....+x = 437.

Answer:

Suppose x is nth term of the given A.P.
an=xHere, a=-4, d=3.It is given that, Sn=437.n22-4+n-13=4373n2-11n-874=03n2-57n+46n-874=03nn-19+46n-19=0n=-463, 19Since, n cannot be in fraction so n=19.Now, an=x-4+19-13=x-4+54=xx=50

Page No 261:

Question 44:

 The sum of first n terms of an A.P. whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another A.P. whose first term is -30 and common difference is 8 . Find n. 

Answer:

According to the question, we have
n228+n-120=2n22-30+2n-1816+20n-20=2-60+16n-820n-4=-136+32n12n=132n=11

Page No 261:

Question 45:

The students of a school decided to beautify the school on the annual day by fixing colourful on the straight passage of the school . They have 27 flags to be fixed at intervals of every 2 metre . The flags are stored at the position of the middle most flag . Ruchi was given the responsibility of placing the flags . Ruchi kept her books where the flags were stored . She could carry only one flag at a time . How much distance did she cover in completing this job and returning back to collect her books ? What is the maximum distance she travelled carrying a flag ?

Answer:

Distance covered to place the first flag to the left of the middle flag = 2 × 2 m = 4 m.
Distance covered to place the second flag to the left of the middle flag = 2 × 4 m = 8 m.
Similarly,
Distance covered to place the thirteenth flag to the left of the middle flag = 2 × 26 m = 52 m.
Now,
The total distance covered = 2 ( 4 + 8 + 12 +....+ 52)
The sum is as follows:
S=2×1324+52=13×56=728 m

The total distance trvelled is 728 m and maximum distance travelled in carrying a flag is 26 m.

Page No 261:

Question 46:

The ratio of the 11th term to the 18th term of an A.P. is 2 : 3. Find the ratio of the 5th term to the 21st term, and also the ratio of the sum of the first five terms to the sum of first 21 terms.

Answer:

Let the first term be a and common difference be d.
Then, 
a11:a18=2:3a+10da+17d=233a+30d=2a+34da=4d                                                      .....1

Now,
a5=a+4d=4d+4d=8da21=a+20d=4d+20d=24da5:a21=8d:24da5:a21=1:3

The sum of the five terms is given by:
S5=522a+5-1d=522×4d+4d=528d+4d=52×12d=30d

Also, the sum of first 21 terms is:
S21=2122a+21-1d=2122×4d+20d=21228d=294d

S5:S21=30d:294d=5:49

Hence, the ratio of the 5th term to the 21st term is 1 : 3 and the ratio of the sum of the first 5 terms to the sum of first 21 terms is 5 : 49.

Page No 261:

Question 1:

Find the common difference of the Arithmetic progression 1a,3-a3a,3-2a3a, ...a0

Answer:

Given: 1a,3-a3a,3-2a3a, ...a0


Common difference= a2-a1                                   =3-a3a-1a                                   =3-a-33a                                   =-a3a                                   =-13

Hence, the common difference of the Arithmetic progression 1a,3-a3a,3-2a3a, ...a0 is -13.

Page No 261:

Question 2:

In an A.P., if the common difference d = – 4, and the seventh term a7 is 4, then find the first term.

Answer:

nth term of an AP is Tn=a+n-1d
For an AP with first term as 'a' and common difference as 'd', the seventh term is a + 6d
According to the question, common difference(d) is −4 and the seventh term(a7) is 4.
So, a + 6will be
⇒ a + 6(−4) = 4
⇒ a = 28
Hence, the first term is 28.

Page No 261:

Question 3:

Write the nth term of the A.P. 1m,1+mm,1+2mm,....

Answer:

Given: A.P. 1m,1+mm,1+2mm,....
We know that the nth term of an AP is given by
an=a+n-1d
In the given AP
a=1md=1+mm-1m=1+m-1m=1
Thus, the nth term of the given AP is
an=1m+n-11=1+n-1mm

Page No 261:

Question 4:

For what value of k will the consecutive terms 2k + 1, 3k + 3 and 5k − 1 form an A.P.?

Answer:

The given terms are 2k + 1, 3k + 3 and 5k − 1. 
The differences between the consecutive terms are
3k + 3 − (2k + 1) = k + 2 = d1                   
and 
5k − 1 − (3k + 3) = 2k − 4 = d2               
If the given terms are in an AP, then
d1= d2
⇒ k + 2 = 2k − 4
⇒ k = 6
Hence, the value of k for which the given terms are in an AP is 6.   

Page No 261:

Question 5:

Find the 9th term from the end of the A.P. 5,  9, 13,...,185.

Answer:

Common difference, d, of the AP = 9 − 5 = 4
Last term, l, of the AP = 185
We know that the nth term from the end of an AP is given by l − (n − 1)d.
Thus, the 9th term from the end is
185 − (9 − 1)4
= 185 − 4 × 8
= 185 − 32
= 153

Page No 261:

Question 6:

If 45, a, 2 are three consecutive terms of an A.P., then find the value of a.

Answer:

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

First term (a1) =

Second term (a2) =

Third term (a3) =

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 262:

Question 7:

For what value of p are 2p + 1, 13, 5p − 3 are three consecutive terms of an A.P.?

Answer:

Here, we are given three terms,

First term (a1) =

Second term (a2) =

Third term (a3) =

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 262:

Question 8:

The first term of an A.P. is p and its common difference is q. Find its 10th term.

Answer:

Here, we are given,

First term (a) = p

Common difference (d) = q

 

We need to find the 10th term (an).

As we know,

So, for 10th term (n = 10), we get,

Therefore,

Page No 262:

Question 9:

Write the value of x for which 2x, x + 10 and 3x + 2 are in A.P.

Answer:

Here, we are given three terms,

First term (a1) =

Second term (a2) =

Third term (a3) =

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

Also,

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

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

Page No 262:

Question 10:

Write 5th term from the end of the A.P. 3, 5, 7, 9, ..., 201.

Answer:

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

3, 5, 7, 9 …201

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

First term (a) = 3

Last term (an) = 201

Common difference (d) =

Now, as we know, the nth term from the end can be given by the formula,

So, the 5th term from the end,

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

Page No 262:

Question 12:

Define an arithmetic progression.

Answer:

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 262:

Question 13:

Which term of the sequence 114, 109, 104, ... is the first negative term?

Answer:

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 262:

Question 14:

Write the value of a30 − a10 for the A.P. 4, 9, 14, 19, ....

Answer:

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

A.P. is

Here,

First term (a) = 4

Common difference of the A.P. (d)

Now, as we know,

Here, we find a30 and a20.

So, for 30th term,

Also, for 10th term,

So,

Therefore, for the given A.P.

Page No 262:

Question 15:

Write the nth term of an A.P. the sum of whose n terms is Sn.

Answer:

We are given an A.P. the sum of whose n terms is Sn. So, to calculate the nth term of the A.P. we use following formula,

So, the nth term of the A.P. is given by .

Page No 262:

Question 16:

Write the sum of first n odd natural numbers.

Answer:

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 262:

Question 17:

Write the sum of first n even natural numbers.

Answer:

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 262:

Question 18:

If the sum of n terms of an A.P. is Sn = 3n2 + 5n. Write its common difference.

Answer:

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 262:

Question 19:

Write the expression of the common difference of an A.P. whose first term is a and nth term is b.

Answer:

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 (an),

Therefore, common difference of the A.P. is

Page No 262:

Question 20:

If the sum of first p term of an A.P. is ap2 + bp, find its common difference.

Answer:

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 262:

Question 21:

Find the sum of the first 10 multiples of 3.

Answer:

Multiples of 3 are:
3, 6, 9, ...

They are in A.P. with
First term (a) = 3
Common difference (d) = 6 − 3 = 3

S10=1022a+10-1d      =52a+9d      =523+93      =56+27      =5×33      =165

​Hence, the sum of the first 10 multiples of 3  is 165.

Page No 262:

Question 22:

How many two digits numbers are divisible by 3?

Answer:

Two digit numbers divisible by 3 are:
12, 15, ..... , 99 

Given:a=12d=15-12  =3an=99Now,an=a+n-1d99=12+n-1399=12+3n-399=9+3n99-9=3n3n=90n=30

Hence, 30 two digits numbers are divisible by 3.

Page No 262:

Question 1:

The sequence 2a-6b3b,2a-3b3b,2a3b,2a+3b3b,2a+6b3b.... is an A.P. with common difference ________

Answer:


The given A.P. is 2a-6b3b,2a-3b3b,2a3b,2a+3b3b,2a+6b3b....

Common difference (d) = a2 − a1
                                       = 2a-3b3b-2a-6b3b
                                       = 2a-3b-2a+6b3b
                                       =​ 3b3b
                                       =​ b

Hence, the sequence 2a-6b3b,2a-3b3b,2a3b,2a+3b3b,2a+6b3b.... is an A.P. with common difference b.

Page No 262:

Question 2:

If the sequence a-3bb,3a-3bb,5a-3bb,7a-3bb,...is an A.P with common difference 3, then a/b = ________.

Answer:


The given A.P. is: a-3bb,3a-3bb,5a-3bb,7a-3bb,...

Common difference (d) = a2 − a1
3=3a-3bb-a-3bb3=3a-3b-a+3bb3=2ab3b=2a2a=3bDividing both sides by 2b, we get2a2b=3b2bab=32
                                       
Hence, if the sequence a-3bb,3a-3bb,5a-3bb,7a-3bb,...is an A.P with common difference 3, then a/b = 32.

Page No 262:

Question 3:

The 11th term of the A.P.-5,-52, 0, 52,..., is

Answer:

The given A.P. is: -5,-52, 0, 52,..., 

First term (a) = –5
Common difference (d) = a2 – a1
                                       =  -52--5
                                       =  -52+5
                                       =  -5+102
                                       =  52

The 11th term is:
a11=a+11-1d     =-5+1052     =-5+55     =-5+25     =20

​Hence, the 11th term of the A.P.-5,-52, 0, 52,..., is 20.

Page No 262:

Question 4:

The 21st term of the A.P. whose first two terms are – 3 and 4 is ____________.

Answer:

Given:
First term (a) = –3
Second term (a2) = 4
Common difference (d) = a2 – a1
                                       =  4--3
                                       =  4 + 3
                                       =  7

The 21st term is:
a21=a+21-1d     =-3+207     =-3+140     =137

​Hence, the 21st term of the A.P. whose first two terms are –3 and 4 is 137.

Page No 262:

Question 5:

The famous Mathematician associated with finding the sum of the first 100 natural numbers was _________.

Answer:

Gauss is the famous Mathematician associated with finding the sum of the first 100 natural numbers.

Hence, the famous Mathematician associated with finding the sum of the first 100 natural numbers was Gauss.

Page No 262:

Question 6:

If n – 2, 4n – 1, and 5n + 2 are in A.P., then n = _______.

Answer:

If – 2, 4n – 1, and 5+ 2 are in A.P.,
then their common difference must be same.

 a2-a1=a3-a2=d4n-1-n-2=5n+2-4n-14n-1-n+2=5n+2-4n+13n+1=n+33n-n=3-12n=2n=1

Hence, = 1.

Page No 262:

Question 7:

The sum of first 50 odd natural numbers is __________.

Answer:

First 50 odd natural numbers are: 1, 3, 5, 7, ...

First term (a) = 1
Common difference (d) = 3 – 1 = 2

Now,
Sn=n22a+n-1dS50=5022a+50-1d      =2521+492      =252+98      =25×100      =2500

Hence, the sum of first 50 odd natural numbers is 2500.

Page No 262:

Question 8:

The sum of first n odd natural numbers is ________.

Answer:

First n odd natural numbers are: 1, 3, 5, 7, ...

First term (a) = 1
Common difference (d) = 3 – 1 = 2

Now,
Sn=n22a+n-1d     =n221+n-12     =n22+2n-2     =n22n     =n2

Hence, the sum of first n odd natural numbers is n2.

Page No 262:

Question 9:

If the common difference of an A.P. is 5, then a18a13 = ______________.

Answer:

Given:
Common difference (d) = 5

Now,
an=a+n-1da18=a+18-1d=a+17da13=a+13-1d=a+12d a18-a13=a+17d-a+12d                     =17d-12d                     =5d                     =5×5        d=5                     =25

Hence, a18 – a13 = 25.

Page No 262:

Question 10:

If a1, a2, a3, ....., an, ..... is an A.P. such that a18a14 = 32, then its common difference is __________.

Answer:

Given:
 a18 – a14 = 32

Now,
an=a+n-1da18=a+18-1d=a+17da14=a+14-1d=a+13d a18-a14=a+17d-a+13d32=17d-13d32=4dd=8

Hence, if a1a2a3, ....., an, ..... is an A.P. such that a18 – a14 = 32, then its common difference is 8.

Page No 262:

Question 11:

If 5, a2, a3, ...., a20, 145 is an A.P., then a2 + a20 =__________.

Answer:

If a1a2a3, ...., an are in A.P., then
a+ a= aan−1 aan−= ....

Here,
aa21 aa20
⇒ aa20 = 5 + 145 = 150

Hence, a2 + a20 = 150.
 

Page No 262:

Question 12:

The value of the middle term of the A.P. –11, –7, –3, ...., 49, 53 is ________.

Answer:

The given A.P. is: –11, –7, –3, ...., 49, 53

First term (a) = –11
Common difference (d) = a2 – a1
                                       =  –7 – (–11)
                                       =  –7 + 11
                                       =  4
Last term (an) = 53

Now,
an=a+n-1d53=-11+n-1453=-11+4n-453=-15+4n4n=53+154n=68n=17

∴ Total number of terms = 17

Middle term=n+12thterm                    =17+!2thterm                    =9th term

Therefore,
a9=a+9-1d    =-11+84    =-11+32    =21

​Hence, the value of the middle term of the A.P. –11, –7, –3, ...., 49, 53 is 21.



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Question 13:

If 9th term of an A.P. is zero, then its 29th and 19th terms are in the ratio __________.

Answer:

Given:
9th term (a9) = 0

Now,
an=a+n-1da9=a+9-1d0=a+8da=-8d

Therefore,
a29a19=a+29-1da+19-1d        =a+28da+18d        =-8d+28d-8d+18d      a=-8d        =20d10d        =21

​Hence, if 9th term of an A.P. is zero, then its 29th and 19th terms are in the ratio 2 : 1.

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Question 14:

If an denotes the nth term of the A.P. 3, 8, 13, 18, ..., then the value of a30 a20 is __________.

Answer:

The given A.P. is: 3, 8, 13, 18, ...

First term  (a) = 3
Common difference (d) = 8 – 3 = 5

Now,
an=a+n-1da30=a+30-1d=a+29da20=a+20-1d=a+19d a30-a20=a+29d-a+19d                     =29d-19d                     =10d                     =10×5        d=5                     =50

Hence, the value of a30 – a20 is 50.

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Question 15:

In an A.P. a1, a2, a3, ..., an, ...., if a1 = 1, an = 20 and Sn = 399, then the value of n is __________.

Answer:

Given:
First term (a) = a1 = 1
nth term = a= 20
Sum of n terms = S= 399

Now,
an=a+n-1d20=1+n-1dn-1d=20-1n-1d=19     ...1     Sn=n22a+n-1d399=n221+19        From 1399=n221798=21nn=79821n=38

Hence, the value of is 38.

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Question 16:

If 7 times the 7th term of an A.P. is equal to 11 times its 11th term, then the value of its 18th term is ________.

Answer:

Given:
7a7=11a11

Now,
an=a+n-1dThus,7a7=11a117a+7-1d=11a+11-1d7a+6d=11a+10d7a+42d=11a+110d11a-7a=42d-110d4a=-68da=-17d       ...1     a18=a+18-1d      =-17d+17d       From 1      =0

Hence, the value of its 18th term is 0.

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Question 17:

Two arithmetic progressions have the same common difference. Their first terms are A and B respectively. The difference between their nth terms is _________.
 

Answer:

Let the first term of first A.P. be A and common difference be d.
And  the first term of second A.P. be B and common difference be d.

 nth term of first A.P. = an
 nth term of second A.P. = bn

Now,
an=A+n-1dand bn=B+n-1dThus,an-bn=A+n-1d-B-n-1d             =A-B

Hence, the difference between their nth terms is A − B.

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Question 18:

If the nth terms of two A.P.s: 9,7,5.... and 24, 21, 18, ... are the same, then the value of n is _________.

Answer:

The given A.P.s are: 
9,7,5.... and 24, 21, 18, ...

9, 7, 5,...
First term (a) = 9
Common difference (d) = 7 − 9 = −2
nth term (an)=a+n-1d                      =9+n-1-2                      =9-2n+2                      =11-2n     ...1

24, 21, 18, ...
First term (a) = 24
Common difference (d) = 21 − 24 = −3
nth term (bn)=a+n-1d                      =24+n-1-3                      =24-3n+3                      =27-3n     ...2

It is given that, an = bn
11-2n=27-3n3n-2n=27-11n=16

Hence, the value of is 16.

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Question 19:

  If Sn = n(4n + 1) is the sum of n terms of an A.P., then its common difference is __________.

Answer:

Given:
Sn=n4n+1Now,Sum of one term=S1=14+1=5a1=a=5      ...1Sum of two terms=S2=28+1=18a1+a2=18       ...2Subtracting 1 from 2, we geta2=13a+d=13d=8       a=5

Hence, if Sn = n(4n + 1) is the sum of n terms of an A.P., then its common difference is 8.

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Question 20:

If the ratio of the sums of first n terms of two A.P.s is 5n+137n+27, then the ratio of their 4th terms is ________.

Answer:

Given:
SnSn'=5n+137n+27Let the first term of 1st A.P. be a and of 2nd A.P. be a'.Let the common difference of 1st A.P. be d and of 2nd A.P. be d'.Let Sn be the sum of n terms of 1st A.P. and Sn' be the sum of n terms of  2nd A.P..Now,SnSn'=n22a+n-1dn22a'+n-1d'5n+137n+27=2a+n-1d2a'+n-1d'            ....1Let a4 be the 4th term of 1st A.P. and a4' be the 4th term of  2nd A.P..a4a4'=a+4-1da'+4-1d'       =a+3da'+3d'Multiplying and dividing the RHS by 2, we geta4a4'=2a+6d2a'+6d'                  ....2Putting n=7 in 1, we get57+1377+27=2a+7-1d2a'+7-1d'35+1349+27=2a+6d2a'+6d'2a+6d2a'+6d'=4876=1219             ....3From 2 and 3, we geta4a4'=1219

Hence, the ratio of their 4th terms is 12 : 19.



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