NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2

# NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2

## NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2 are the part of NCERT Solutions for Class 11 Maths. Here you can find the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2.

### Ex 9.2 Class 11 Maths Question 1.

Find the sum of odd integers from 1 to 2001.

Solution:
We have to find 1 + 3 + 5 + ……….. + 2001.
This series is in A.P. with first term a = 1, common difference d = 3 – 1 = 2 and last term l = 2001.
l = a + (n – 1)d

2001 = 1 + (n – 1)2

2001 = 1 + 2n – 2

2n = 2001 + 1

### Ex 9.2 Class 11 Maths Question 2.

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

Solution:

The multiples of 5 lying between 100 and 1000 are 105, 110, 115, …, 995.
We have to find 105 + 110 + 115 + …….. + 995
This series is in A.P. with first term a = 105, common difference d = 110 – 105 = 5 and last term l = 995.

### Ex 9.2 Class 11 Maths Question 3.

In an A.P., the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is -112.

Solution:
Given, the first term a = 2. Let d be the common difference.

### Ex 9.2 Class 11 Maths Question 4.

How many terms of the A.P.  -6, 11/2,  -5, ……. are needed to give the sum –25?

Solution:
Let a be the first term and d be the common difference of the given A.P., then

### Ex 9.2 Class 11 Maths Question 5.

In an A.P., if pth term is 1/q and qth term is 1/p, prove that the sum of first pq terms is ½ (pq + 1), where p ≠ q.

Solution:
Let a be the first term and d be the common difference of the given A.P., then

### Ex 9.2 Class 11 Maths Question 6.

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

Solution:
Let a be the first term and d be the common difference.
We have a = 25, d = 22 – 25 = -3, Sn = 116

### Ex 9.2 Class 11 Maths Question 7.

Find the sum to n terms of the A.P., whose kth term is 5k + 1.

Solution:
Given, ak = 5k + 1
By substituting the value of k = 1, 2, 3 and 4, we get

### Ex 9.2 Class 11 Maths Question 8.

If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants, find the common difference.

Solution:
Given, Sn = pn + qn2, where Sn be the sum of n terms.

### Ex 9.2 Class 11 Maths Question 9.

The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18th terms.

Solution:
Let a1, a2 & d1, d2 be the first terms & common differences of the two arithmetic progressions, respectively. According to the given condition, we have

### Ex 9.2 Class 11 Maths Question 10.

If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.

Solution:
Let the first term be a and the common difference be d.
According to the question,

### Ex 9.2 Class 11 Maths Question 11.

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

Solution:
Let the first term be A & common difference be D. We have

### Ex 9.2 Class 11 Maths Question 12.

The ratio of the sums of m and n terms of an A.P. is m2 : n2. Show that the ratio of mth and nth term is (2m – 1) : (2n – 1).

Solution:
Let the first term be a & the common difference be d. We have,

### Ex 9.2 Class 11 Maths Question 13.

If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m.

Solution:
Given, Sn = 3n2 + 5n, where Sn be the sum of n terms.

### Ex 9.2 Class 11 Maths Question 14.

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

Solution:
Let A1, A2, A3, A4, A5 be the numbers between 8 and 26 such that 8, A1, A2, A3, A4, A5, 26 are in A.P. Here a = 8, l = 26, n = 7

### Ex 9.2 Class 11 Maths Question 15.

If (an + bn)/(an1 + bn1) is the A.M. between a and b, then find the value of n.

Solution:
We have, (an + bn)/(an1 + bn1) = (a + b)/2

### Ex 9.2 Class 11 Maths Question 16.

Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7thand (m – 1)th numbers is 5 : 9. Find the value of m.

Solution:

Let A1, A2, ………, Am be m numbers which have been inserted between 1 and 31.
Then, the sequence be 1, A1, A2, ……… Am, 31 is in A.P.

Here, a = 1, 31 is (m + 2)th term, let d be the common difference.

### Ex 9.2 Class 11 Maths Question 17.

A man starts repaying a loan as first instalment of Rs. 100. If he increases the instalment by Rs. 5 every month, what amount he will pay in the 30th instalment?

Solution:

The first, second, third, … instalments are Rs 100, Rs 105, Rs 110, …
Here, we have an A.P. with a = 100 and d = 5.
We have, an = a + (n – 1)d

a30 = a + 29d = 100 + 29(5) = 100 + 145 = 245
Hence, the man will pay Rs. 245 in 30th instalment.

### Ex 9.2 Class 11 Maths Question 18.

The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.

Solution:
Let there are n sides of a polygon.

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