NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.2

NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.2: NCERT Solutions for Class 11 Maths Chapter 8, Exercise 8.2 focus on Geometric Progressions (GPs) in Sequences and Series. A Geometric Progression is a sequence where each term is obtained by multiplying the previous term by a fixed number called the common ratio (r).

 This exercise helps students understand patterns, exponential growth, and practical applications in finance, physics, and real-world problem-solving.

NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.2 Overview

NCERT Solutions for Class 11 Maths Chapter 8, Exercise 8.2 focus on Geometric Progressions (GPs), an essential concept in Sequences and Series. A GP is a sequence where each term is obtained by multiplying the previous term by a fixed common ratio (r). 

Understanding GPs is important in real-world applications like compound interest, population growth, physics, and financial calculations, making this exercise highly valuable for students.

NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.2 PDF

Below, we have provided the NCERT Solutions for Class 11 Maths Chapter 8, Exercise 8.2 in PDF format. This exercise focuses on Geometric Progressions (GPs), covering key concepts like the nth term and sum of n terms. These solutions offer step-by-step explanations to help students understand and solve problems efficiently. Click the link below to download the PDF and strengthen your grasp of Sequences and Series.

NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.2 PDF

NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.2 Sequences and Series

Below is the NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.2 Sequences and Series -

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,

Sn = 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,

Sn = 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 20th term is –112.

Solution:

Given,

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

Let’s assume d is 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 20th 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 known that,

Sn = 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 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:

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 is n.

So, we have

Sn = 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 8th 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 kth term is 5k + 1.

Solution:

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

kth term = ak = 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,

Sn = 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 18th 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 the mth and the nth term is (2m – 1): (2n – 1).

Solution:

Let’s consider that a and b are 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 mth 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.

am = a + (m – 1)d = 164 … (1)

The sum of the terms is given by,

Sn = 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, 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 7th and (m – 1)th numbers is 5: 9. Find the value of m.

Solution:

Let’s consider a1, a2, … am be m numbers such that 1, a1, a2, … am, 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

am–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 30th 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 then, A.P. is 100, 105, 110, …

Where the first term, a = 100

Common difference, d = 5

So, the 30th 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 30th 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.

Benefits of Using NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.2 Sequences and Series

Comprehensive Understanding of Geometric Progressions (GPs)

• This exercise focuses on Geometric Progressions (GPs), helping students understand the concept of common ratio, nth term, and sum of terms systematically.

Step-by-Step Problem Solving

• Solutions are presented in a structured manner, breaking down each step logically, making it easier for students to grasp concepts.

Strengthens Conceptual Knowledge

• Helps students understand patterns in number sequences, which is crucial for solving real-world problems in fields like finance, physics, and data science.

Exam-Oriented Approach

• NCERT solutions follow the CBSE exam pattern, ensuring students get well-structured answers as expected in board exams.

Error-Free and Reliable Solutions

• Solutions are prepared by subject experts, ensuring accuracy and clarity, which helps avoid common mistakes.

Saves Time in Exam Preparation

• Provides quick and easy reference for students to revise efficiently before exams, making it a time-saving tool.

Boosts Problem-Solving and Logical Thinking

• Improves students' analytical skills by teaching them to recognize patterns and sequences, which is useful for higher mathematics.