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

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 8.

Chapter 9 Sequences and Series of Class 11 Maths is categorised under the CBSE Syllabus for 2023-24. The second exercise of this chapter is based on the topic of Arithmetic Progression and Arithmetic Mean. Arithmetic progression (A.P.) or arithmetic sequence is a sequence of numbers such that the difference between consecutive terms is constant. The exercise also deals with questions related to Arithmetic Mean, which is defined as the average of a set of numerical values, as calculated by adding them together and dividing by the number of terms in the set. Learn and understand more about these topics by solving Exercise 9.2 of NCERT Solutions for Class 11 Maths Chapter 9 – Sequences and Series.

There are a lot of questions given in the NCERT textbook for the students to solve and practise. Solving the NCERT Solutions for Class 11 Maths and practising will undoubtedly aid the students in scoring high marks in the Class 11 examinations. Students have to ensure that they practise every problem given in the textbook repeatedly until the concept gets clear.

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

Download PDF

carouselExampleControls112

Previous





Next

Access Other Exercise Solutions for NCERT Class 11 Maths Chapter 9 – Sequences and Series

Exercise 9.1 Solutions 14 Questions

Exercise 9.3 Solutions 32 Questions

Exercise 9.4 Solutions 10 Questions

Miscellaneous Exercise on Chapter 9 Solutions 32 Questions

Also explore – NCERT Class 11 Solutions

Access Solutions for NCERT Class 11 Maths Chapter 9 Exercise 9.2

1. Find the sum of odd integers from 1 to 2001.

Solution:

The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001.

It clearly forms a sequence in A.P.

Where, the first term, a = 1

The common difference, d = 2

Now,

a + (n -1)d = 2001

1 + (n-1)(2) = 2001

2n – 2 = 2000

2n = 2000 + 2 = 2002

n = 1001

We know,

S_n = n/2 [2a + (n-1)d]

Therefore, the sum of odd numbers from 1 to 2001 is 1002001.

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

Solution:

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.

It clearly forms a sequence in A.P.

Where, the first term, a = 105

The common difference, d = 5

Now,

a + (n -1)d = 995

105 + (n – 1)(5) = 995

105 + 5n – 5 = 995

5n = 995 – 105 + 5 = 895

n = 895/5

n = 179

We know,

S_n = n/2 [2a + (n-1)d]

Therefore, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.

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 the 20^th term is –112.

Solution:

Given,

The first term (a) of an A.P. = 2

Let’s assume d be the common difference of the A.P.

So, the A.P. will be 2, 2 + d, 2 + 2d, 2 + 3d, …

Then,

Sum of first five terms = 10 + 10d

Sum of next five terms = 10 + 35d

From the question, we have

10 + 10d = ¼ (10 + 35d)

40 + 40d = 10 + 35d

30 = -5d

d = -6

a₂₀ = a + (20 – 1)d = 2 + (19) (-6) = 2 – 114 = -112

Therefore, the 20^th term of the A.P. is –112.

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

Solution:

Let’s consider the sum of n terms of the given A.P. as –25.

We know that,

S_n = n/2 [2a + (n-1)d]

where n = number of terms, a = first term, and d = common difference

So here, a = –6

d = -11/2 + 6 = (-11 + 12)/ 2 = 1/2

Thus, we have

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

Solution:

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

Solution:

Given A.P.,

25, 22, 19, …

Here,

First term, a = 25 and

Common difference, d = 22 – 25 = -3

Also given, the sum of a certain number of terms of the A.P. is 116

The number of terms be n

So, we have

S_n = n/2 [2a + (n-1)d] = 116

116 = n/2 [2(25) + (n-1)(-3)]

116 x 2 = n [50 – 3n + 3]

232 = n [53 – 3n]

232 = 53n – 3n²

3n² – 53n + 232 = 0

3n² – 24n – 29n+ 232 = 0

3n(n – 8) – 29(n – 8) = 0

(3n – 29) (n – 8) = 0

Hence,

n = 29/3 or n = 8

As n can only be an integral value, n = 8

Thus, the 8^th term is the last term of the A.P.

a₈ = 25 + (8 – 1)(-3)

= 25 – 21

= 4

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

Solution:

Given, the k^th term of the A.P. is 5k + 1.

k^th term = a_k = a + (k – 1)d

And,

a + (k – 1)d = 5k + 1

a + kd – d = 5k + 1

On comparing the coefficient of k, we get d = 5

a – d = 1

a – 5 = 1

⇒ a = 6

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 know that,

S_n = n/2 [2a + (n-1)d]

From the question we have,

On comparing the coefficients of n² on both sides, we get

d/2 = q

Hence, d = 2q

Therefore, the common difference of the A.P. is 2q.

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, and d₁, d₂ be the first terms and the common difference of the first and second arithmetic progression, respectively.

Then, from the question we have

10. If the sum of the 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’s take a and d to be the first term and the common difference of the A.P., respectively.

Then, it is given that

Therefore, the sum of (p + q) terms of the A.P. is 0.

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

Solution:

Let a₁ and d be the first term and the common difference of the A.P., respectively.

Then according to the question, we have

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

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’s consider a and b to be the first term and the common difference of the A.P., respectively.

Then from the question, we have

Hence, the given result is proved.

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

Solution:

Let’s consider a and b to be the first term and the common difference of the A.P., respectively.

a_m = a + (m – 1)d = 164 … (1)

The sum of the terms is given by,

S_n = n/2 [2a + (n-1)d]

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

Solution:

Let’s assume A₁, A₂, A₃, A₄, and A₅ to be five numbers between 8 and 26 such that 8, A₁, A₂, A₃, A₄, A₅, and 26 are in an A.P.

Here we have,

a = 8, b = 26, n = 7

So,

26 = 8 + (7 – 1) d

6d = 26 – 8 = 18

d = 3

Now,

A₁ = a + d = 8 + 3 = 11

A₂ = a + 2d = 8 + 2 × 3 = 8 + 6 = 14

A₃ = a + 3d = 8 + 3 × 3 = 8 + 9 = 17

A₄ = a + 4d = 8 + 4 × 3 = 8 + 12 = 20

A₅ = a + 5d = 8 + 5 × 3 = 8 + 15 = 23

Therefore, the required five numbers between 8 and 26 are 11, 14, 17, 20, and 23.

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

Solution:

The A.M. between a and b is given by, (a + b)/2

Then according to the question,

Thus, the value of n is 1.

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^th and (m – 1)^th numbers is 5:9. Find the value of m.

Solution:

Let’s consider a₁, a₂, … a_m be m numbers such that 1, a₁, a₂, … a_m, 31 is an A.P.

And here,

a = 1, b = 31, n = m + 2

So, 31 = 1 + (m + 2 – 1) (d)

30 = (m + 1) d

d = 30/ (m + 1) ……. (1)

Now,

a₁ = a + d

a₂ = a + 2d

a₃ = a + 3d …

Hence, a₇ = a + 7d

a_m–1 = a + (m – 1) d

According to the question, we have

Therefore, the value of m is 14.

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

Solution:

Given,

The first instalment of the loan is Rs. 100.

The second instalment of the loan is Rs. 105, and so on, as the instalment increases by Rs. 5 every month.

Thus, the amount that the man repays every month forms an A.P.

And the A.P. is 100, 105, 110, …

Where, the first term, a = 100

The common difference, d = 5

So, the 30^th term in this A.P. will be

A₃₀  = a + (30 – 1)d

= 100 + (29) (5)

= 100 + 145

= 245

Therefore, the amount to be paid in the 30^th instalment will be Rs. 245.

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:

It’s understood from the question that the angles of the polygon will form an A.P. with a common difference d = 5° and first term a = 120°.

And, we know that the sum of all angles of a polygon with n sides is 180° (n – 2).

Thus, we can say

Thus, a polygon having 9 and 16 sides will satisfy the condition in the question.