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 part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2.

• Sequences and Series Class 11 Ex 9.1
• Sequences and Series Class 11 Ex 9.3
• Sequences and Series Class 11 Ex 9.4

Board	CBSE
Textbook	NCERT
Class	Class 11
Subject	Maths
Chapter	Chapter 9
Chapter Name	Sequences and Series
Exercise	Ex 9.2
Number of Questions Solved	18
Category	NCERT Solutions

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

Question 1.
Find the sum of odd integers from 1 to 2001.
Solution:
We have to find 1 + 3 + 5 + ……….. + 2001
This is an 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

Question 2.
Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
Solution:
We have to find 105 +110 +115 + ……..+ 995
This is an A.P. with first term a = 105, common difference d = 110 -105 = 5 and last term 1 = 995

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:
Let a = 2 be the first term and d be the common difference.

Question 4.
How many terms of the A.P. -6, , -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., we have

Question 5.
In an A.P., if pth term is and qth term is prove that the sum of first pq terms is , where p±q.
Solution:
Let a be the first term & d be the common difference of the A.P., then

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, S_n = 116

Question 7.
Find the sum to n terms of the A.P., whose k^th term is 5k + 1.
Solution:
We have a_k = 5k +1
By substituting the value of k = 1, 2, 3 and 4,
we get

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:
We have S_n = pn + qn², where S„ be the sum of n terms.

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

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 common difference be d.
According to question

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

Question 12.
The ratio of the sums of m and n terms of an A.P. is m²: n². Show that the ratio of m^th and n^th term is (2m -1): (2n -1).
Solution:
Let the first term be a & common difference be d. Then

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:
We have S_n = 3n² + 5n, where Sn be the sum of n terms.

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 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

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

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 7^thand (m – 1 )^th numbers is 5:9. Find the value of m.
Solution:
Let the sequence be 1, A₁, A₂, ……… A_m, 31 Then 31 is (m + 2)^th term, a = 1, let d be the common difference

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:
Here, we have an A.P. with a = 100 and d = 5
∴ a_n= a + 29d = 100 + 29(5) = 100 + 145 = 245
Hence he will pay Rs. 245 in 30^th instalment.

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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