**NCERT solutions for class 11 maths chapter 9 Sequences and Series: **Sequence means the progression of numbers in a definite order and series means the sum of the objects of the sequence. In the previous classes, you have studied about arithmetic progression(A.P). In this chapter, we will discuss more arithmetic progression(A.P) and geometric progression(G.P). In this article, you will get NCERT solutions for class 11 maths chapter 9 sequences and series. Important topics like arithmetic progression(A.P), geometric progression(G.P), arithmetic means(A.M), geometric mean(G.M), the relationship between A.M. and G.M, sum to n terms of special series, sum to n terms of squares and cubes of natural numbers are covered in this chapter. You will get questions related to these topics in the solutions of NCERT for class 11 maths chapter 9 sequences and series. In this chapter, there are two types of sequence.

- Finite sequence( A sequence containing a finite number of terms)
- Infinite sequence( A sequence has a first term but doesn't have last term or a sequence which is not finite)

Check all **NCERT solutions** from class 6 to 12 to understand the concepts in a much easy way.

9.1 Introduction

9.2 Sequences

9.3 Series

9.4 Arithmetic Progression (A.P.)

9.5 Geometric Progression (G.P.)

9.6 Relationship Between A.M. and G.M.

9.7 Sum to n terms of Special Series

**Question:2** Write the first five terms of each of the sequences in Exercises 1 to 6 whose nthterms are:

**Answer:**

Given :

Therefore, the required number of terms

**Question:3** Write the first five terms of each of the sequences in Exercises 1 to 6 whose nthterms are:

**Answer:**

Given :

Therefore, required number of terms

**Question:4** Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

**Answer:**

Given :

Therefore, the required number of terms

**Question:5 **Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

**Answer:**

Given :

Therefore, the required number of terms

**Question:6 **Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

**Answer:**

Given :

Therefore, the required number of terms

**Question:7** Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

**Answer:**

Put

Put n=24,

Hence, we have

**Question:8 **Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

**Answer:**

Given :

Put n=7,

Heence, we have

**Question:9 **Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

**Answer:**

Given :

Put n =9,

The value of

**Question:10 **Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

**Answer:**

Given :

Put n=20,

Hence, value of

**Answer:**

Given :

Hence, five terms of series are

Series

**Answer:**

Given :

Hence, five terms of series are

Series

**Answer:**

Given :

Hence, five terms of series are

Series

**Question:14 **The Fibonacci sequence is defined by

**Answer:**

Given : The Fibonacci sequence is defined by

**Solutions of NCERT for class 11 maths chapter 9 sequences and series-Exercise: 9.2**

**Question:1** Find the sum of odd integers from 1 to 2001.

**Answer:**

Odd integers from 1 to 2001 are

This sequence is an A.P.

Here , first term =a =1

common difference = 2.

We know ,

The , sum of odd integers from 1 to 2001 is 1002001.

**Question:2** Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

**Answer:**

**Numbers divisible by 5** from 100 to 1000 are

This sequence is an A.P.

Here , first term =a =105

common difference = 5.

We know ,

The sum of numbers divisible by 5 from 100 to 1000 is 98450.

**Answer:**

First term =a=2

Let the series be

Sum of first five terms

Sum of next five terms

Given : The sum of the first five terms is one-fourth of the next five terms.

To prove :

L.H.S :

Hence, 20th term is –112.

**Question:****4** How many terms of the A.P. are needed to give the sum –25?

**Answer:**

Given : A.P. =

Given : sum = -25

**Answer:**

Given : In an A.P., if pth term is 1/q and qth term is 1/p

Subtracting (2) from (1), we get

Putting value of d in equation (1),we get

Hence,the sum of first pq terms is 1/2 (pq +1), where .

**Question:6 **If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term.

**Answer:**

Given : A.P. 25, 22, 19, ….....

a=25 , d = -3

n could not be so n=8.

Last term

The, last term of A.P. is 4.

**Question:7** Find the sum to n terms of the A.P., whose term is 5k + 1.

**Answer:**

Given :

Comparing LHS and RHS , we have

and

Putting value of d,

**Question:8** If the sum of n terms of an A.P. is , where p and q are constants, find the common difference

**Answer:**

If the sum of n terms of an A.P. is ,

Comparing coefficients of on both side , we get

The common difference of AP is 2q.

**Answer:**

Given: The sums of n terms of two arithmetic progressions are in the ratio.

There are two AP's with first terms = and common difference =

Substituting n=35,we get

Thus, the ratio of the 18th term of AP's is

**Answer:**

Let first term of AP = a and common difference = d.

Then,

Given :

Now,

Thus, sum of p+q terms of AP is 0.

**Question:11 **Sum of the first p, q and r terms of an A.P. are a, b and c, respectively. Prove that

**Answer:**

To prove :

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

According to the given information, we have

Subtracting equation (2) from (1), we have

Subtracting equation (3) from (2), we have

Equating values of d, we have

Dividing both sides from pqr, we get

Hence, the given result is proved.

**Question:12** The ratio of the sums of m and n terms of an A.P. is . Show that the ratio of mth and nth term is .

**Answer:**

Let a and b be the first term and common difference of a AP ,respectively.

Given : The ratio of the sums of m and n terms of an A.P. is .

To prove : the ratio of mth and nth term is .

Put , we get

From equation (1) ,we get

Hence proved.

**Question:13** If the sum of n terms of an A.P. is and its term is 164, find the value of m.

**Answer:**

Given : If the sum of n terms of an A.P. is and its term is 164

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

Sum of n terms =

Comparing the coefficients of n on both side , we have

Also ,

m th term is 164.

Hence, the value of m is 27.

**Question:14** Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

**Answer:**

Let five numbers be A,B,C,D,E.

Then

Here we have,

Thus, we have

Thus, the five numbers are 11,14,17,20,23.

**Question:15** If is the A.M. between a and b, then find the value of n.

**Answer:**

Given : is the A.M. between a and b.

Thus, value of n is 1.

**Answer:**

Let A,B,C.........M be m numbers.

Then,

Here we have,

Given : the ratio of and numbers is 5 : 9.

Putting value of d from above,

Thus, value of m is 14.

**Answer:**

The first instalment is of Rs. 100.

If the instalment increase by Rs 5 every month, second instalment is Rs.105.

Then , it forms an AP.

We have ,

Thus, he will pay Rs. 245 in the 30th instalment.

**Answer:**

The angles of polygon forms AP with common difference of and first term as .

We know that sum of angles of polygon with n sides is

Sides of polygon are 9 or 16.

**CBSE NCERT solutions for class 11 maths chapter 9 sequences and series-Exercise: 9.3**

**Question:1** Find the and terms of the G.P.

**Answer:**

G.P :

first term = a

common ratio =r

the nth term of G.P

**Question:2** Find the term of a G.P. whose term is 192 and the common ratio is 2.

**Answer:**

First term = a

common ratio =r=2

term is 192

is the term of a G.P.

**Question:3 **The terms of a G.P. are p, q and s, respectively. Show that

**Answer:**

To prove :

Let first term=a and common ratio = r

Dividing equation 2 by 1, we have

Dividing equation 3 by 2, we have

Equating values of , we have

Hence proved

**Question:4 **The term of a G.P. is square of its second term, and the first term is -3. Determine its term.

**Answer:**

First term =a= -3

term of a G.P. is square of its second term

Thus, seventh term is -2187.

**Question:5(a) **Which term of the following sequences:

**Answer:**

Given :

n th term is given as 128.

The, 13 th term is 128.

**Question:5(b)** Which term of the following sequences:

**Answer:**

Given :

n th term is given as 729.

The, 12 th term is 729.

**Question:5(c) **Which term of the following sequences:

**Answer:**

Given :

n th term is given as

Thus, n=9.

**Question:6** For what values of x, the numbers are in G.P.?

**Answer:**

Common ratio=r.

Thus, for ,given numbers will be in GP.

**Answer:**

geometric progressions is 0.15, 0.015, 0.0015, ... .....

a=0.15 , r = 0.1 , n=20

**Question:9 **Find the sum to indicated number of terms in each of the geometric progressions in

**Answer:**

The sum to the indicated number of terms in each of the geometric progressions is:

**Question:12** The sum of first three terms of a G.P. is and their product is 1. Find the common ratio and the terms.

**Answer:**

Given : The sum of first three terms of a G.P. is and their product is 1.

Let three terms be .

Product of 3 terms is 1.

Put value of a in equation 1,

The three terms of AP are .

**Question:13** How many terms of G.P. , … are needed to give the sum 120?

**Answer:**

G.P.= , …............

Sum =120

These terms are GP with a=3 and r=3.

Hence, we have value of n as 4 to get sum of 120.

**Answer:**

Let GP be

Given : The sum of first three terms of a G.P. is 16

Given : the sum of the next three terms is128.

Dividing equation (2) by (1), we have

Putting value of r =2 in equation 1,we have

**Question:15** Given a G.P. with a = 729 and term 64, determine

**Answer:**

Given a G.P. with a = 729 and term 64.

**(Answer)**

**Question:16** Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term

**Answer:**

Given : sum of the first two terms is – 4 and the fifth term is 4 times the third term

Let first term be a and common ratio be r

If r=2, then

If r= - 2, then

Thus, required GP is or

**Question:17** If the terms of a G.P. are x, y and z, respectively. Prove that x,y, z are in G.P.

**Answer:**

Let x,y, z are in G.P.

Let first term=a and common ratio = r

Dividing equation 2 by 1, we have

Dividing equation 3 by 2, we have

Equating values of , we have

Thus, x,y,z are in GP

**Question:18** Find the sum to n terms of the sequence, 8, 88, 888, 8888… .

**Answer:**

8, 88, 888, 8888… is not a GP.

It can be changed in GP by writing terms as

to n terms

**Answer:**

Here, is a GP.

first term =a=4

common ratio =r

**Question:20 **Show that the products of the corresponding terms of the sequences form a G.P, and find the common ratio.

**Answer:**

To prove : is a GP.

Thus, the above sequence is a GP with common ratio of rR.

**Answer:**

Let first term be a and common ratio be r.

Given : the third term is greater than the first term by 9, and the second term is greater than the by 18.

Dividing equation 2 by 1 , we get

Putting value of r , we get

Thus, four terms of GP are

**Question:22** If the terms of a G.P. are a, b and c, respectively. Prove that

**Answer:**

To prove :

Let A be the first term and R be common ratio.

According to the given information, we have

L.H.S :

=RHS

Thus, LHS = RHS.

Hence proved.

**Answer:**

Given : First term =a and n th term = b.

Common ratio = r.

To prove :

Then ,

P = product of n terms

Here, is a AP.

Put in equation (2),

Hence proved .

**Question:24** Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from term is

**Answer:**

Let first term =a and common ratio = r.

Since there are n terms from (n+1) to 2n term.

Sum of terms from (n+1) to 2n.

Thus, the required ratio =

Thus, the common ratio of the sum of first n terms of a G.P. to the sum of terms from term is .

**Question:25** If a, b, c and d are in G.P. show that

**Answer:**

If a, b, c and d are in G.P.

To prove :

RHS :

Using equation (1) and (2),

= LHS

Hence proved

**Question:26** Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

**Answer:**

Let A, B be two numbers between 3 and 81 such that series 3, A, B,81 forms a GP.

Let a=first term and common ratio =r.

For ,

The, required numbers are 9,27.

**Question:27** Find the value of n so that may be the geometric mean between a and b.

**Answer:**

M of a and b is

Given :

Squaring both sides ,

**Question:28 **The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio

**Answer:**

Let there be two numbers a and b

geometric mean

According to the given condition,

.............................................................(1)

Also,

.......................................................(2)

From (1) and (2), we get

Putting the value of 'a' in (1),

Thus, the ratio is

**Question:29** If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are

**Answer:**

If A and G be A.M. and G.M., respectively between two positive numbers,

Two numbers be a and b.

...................................................................1

...........................................................................2

We know

Put values from equation 1 and 2,

..................................................................3

From 1 and 3 , we have

Put value of a in equation 1, we get

Thus, numbers are

**Answer:**

The number of bacteria in a certain culture doubles every hour.It forms GP.

Given : a=30 and r=2.

Thus, bacteria present at the end of the 2nd hour, 4th hour and nth hour are 120,480 and respectively.

**Answer:**

Given: Bank pays an annual interest rate of 10% compounded annually.

Rs 500 amounts are deposited in the bank.

At the end of the first year, the amount

At the end of the second year, the amount

At the end of the third year, the amount

At the end of 10 years, the amount

Thus, at the end of 10 years, amount

**Answer:**

Let roots of the quadratic equation be a and b.

According to given condition,

We know that

Thus, the quadratic equation =

**NCERT solutions for class 11 maths chapter 9 sequences and series-Exercise: 9.4**

**Question:1** Find the sum to n terms of each of the series in

**Answer:**

the series =

n th term =

**Question:2 **Find the sum to n terms of each of the series in

**Answer:**

the series =

n th term =

Thus, sum is

**Question:3** Find the sum to n terms of each of the series

**Answer:**

the series

nth term =

Thus, the sum is

**Question:4 **Find the sum to n terms of each of the series in

**Answer:**

Series =

.................................

Hence, the sum is

**Question:5 **Find the sum to n terms of each of the series in

**Answer:**

series =

n th term =

16th term is

Hence, the sum of the series is 2840.

**Question:6** Find the sum to n terms of each of the series

**Answer:**

series =

=(n th term of 3,6,9,...........)(nth terms of 8,11,14,..........)

n th term =

Hence, sum is

**Question:8** Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by

**Answer:**

nth terms is given by

**Question:9** Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by

**Answer:**

nth terms are given by

This term is a GP with first term =a =2 and common ratio =r =2.

Thus, the sum is

**Question:10** Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by

**Answer:**

nth terms is given by .

**Question:1 **Show that the sum of and terms of an A.P. is equal to twice the term.

**Answer:**

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

Kth term of a AP is given by,

Hence, the sum of and terms of an A.P. is equal to twice the term.

**Question:2** If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.

**Answer:**

Let three numbers of AP are a-d, a, a+d.

According to given information ,

When d=3, AP= 5,8,11 also if d=-3 ,AP =11,8,5.

Thus, three numbers are 5,8,11.

**Question:3 **Let the sum of n, 2n, 3n terms of an A.P. be , respectively, show that

**Answer:**

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

Subtract equation 1 from 2,

Hence, the result is proved.

**Question:4 **Find the sum of all numbers between 200 and 400 which are divisible by 7.

**Answer:**

**Numbers divisible by 7** from 200 to 400 are

This sequence is an A.P.

Here , first term =a =203

common difference = 7.

We know ,

The sum of numbers divisible by 7from 200 to 400 is 8729.

**Question:5** Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

**Answer:**

**Numbers divisible by 2 ** from 1 to 100 are

This sequence is an A.P.

Here , first term =a =2

common difference = 2.

We know ,

**Numbers divisible by 5 ** from 1 to 100 are

This sequence is an A.P.

Here , first term =a =5

common difference = 5.

We know ,