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
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 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
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:
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
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:
Let a = 2 be the first term and d be the common difference.
Ex 9.2 Class 11 Maths Question 4.
How many terms of the A.P. -6, \(-\frac { 11 }{ 2 } \), -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., we have
Ex 9.2 Class 11 Maths Question 5.
In an A.P., if pth term is \(\frac { 1 }{ q } \) and qth term is \(\frac { 1 }{ p } \) prove that the sum of first pq terms is \(\frac { 1 }{ 2 } \left( pq+1 \right) \), where p±q.
Solution:
Let a be the first term & d be the common difference of the 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:
We have 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:
We have Sn = pn + qn2, where S„ 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 common difference be d.
According to 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 & common difference be d. Then
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:
We have 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 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 \(\frac { { a }^{ n }+{ b }^{ n } }{ { a }^{ n-1 }+{ b }^{ n-1 } } \) is the A.M. between a and b, then find the value of n.
Solution:
We have \(\frac { { a }^{ n }+{ b }^{ n } }{ { a }^{ n-1 }+{ b }^{ n-1 } } =\frac { a+b }{ 2 } \)
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 the sequence be 1, A1, A2, ……… Am, 31 Then 31 is (m + 2)th term, a = 1, 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:
Here, we have an A.P. with a = 100 and d = 5
∴ an= a + 29d = 100 + 29(5) = 100 + 145 = 245
Hence he 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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