Download RD Sharma Solutions for Class 11 Chapter 20 – Geometric Progressions

RD Sharma Solutions for Class 11 Chapter 20 – Geometric Progressions is an important resource for students who want to understand the concept of sequences and series in maths. Geometric Progressions (G.P.) are used in many real-life situations, like calculating compound interest, studying population growth, or observing patterns in nature. If you’re looking for simple explanations and step-by-step answers, this chapter makes learning smooth and easy.

In the Geometric Progressions Chapter, students first learn what a geometric progression is—a sequence where each term is found by multiplying the previous term by a fixed, non-zero number called the common ratio. For example, the sequence 3, 6, 12, 24, ... has a common ratio of 2. Once this basic idea is clear, students can move on to more advanced problems.

Next, the chapter introduces important formulas like the nth term of a G.P. and the sum of n terms. It also shows how to use these formulas in different types of questions. Students learn about infinite geometric series, geometric means, and how to insert geometric means between numbers—topics that are useful for both higher studies and competitive exams.

Do Check: RD Sharma Solutions for Class 6 to 12

RD Sharma Solutions for Class 11 Chapter 20 give step-by-step answers to all textbook questions. The language is simple, and each step is explained clearly so students can easily follow the method. The RD Sharma Solutions Class 11 also include diagrams, solved examples, and tips to help boost exam preparation.

By studying RD Sharma Solutions for Class 11, students not only score better in exams but also build a strong understanding of the topic. Whether students are preparing for a JEE or want to improve your concepts in geometric progressions, RD Sharma Chapter 20 Solutions are your perfect guide to success.

Do Check: RD Sharma Solutions for Class 11 Arithmetic Progression

Download RD Sharma Solutions for Class 11 Chapter 20 – Geometric Progressions PDF Here

The RD Sharma Class 11 Chapter 20 PDF contains detailed solutions, solved examples, and extra questions to help you understand geometric progressions better.

Access Answers to RD Sharma Solutions for Class 11 Chapter 20 – Geometric Progressions

This chapter explains geometric progressions, their properties, and how to solve related problems. The RD Sharma Solutions for Class 11 are easy to follow and help students understand each concept better so they can apply it with confidence.

RD Sharma Solutions for Class 11 Geometric Progressions Exercises 

Question 1: Define a geometric progression and write the first five terms of a G.P. whose first term is 3 and the common ratio is 2.

Solution:

A geometric progression (G.P.) is a sequence where each term is obtained by multiplying the previous term by a fixed number called the common ratio.

Given: First term (a) = 3, Common ratio (r) = 2

The general term of a G.P. is: an = a × r(n-1)

First five terms:

• 1st term = 3
• 2nd term = 3 × 2 = 6
• 3rd term = 3 × 2² = 12
• 4th term = 3 × 2³ = 24
• 5th term = 3 × 2⁴ = 48

Question 2: Problem: Find the 12th term of the G.P. 5, 10, 20, 40, ...

Solution:

Given G.P.: 5, 10, 20, 40, ...

First term (a) = 5

Common ratio (r) = 10/5 = 2

Using the formula: an = a × r(n-1)

12th term = 5 × 2(12-1)

= 5 × 2¹¹

= 5 × 2048

= 10,240

Question 3: If the 7th term of a G.P. is 192 and the common ratio is 2, find the first term.

Solution:

Given: 7th term = 192, Common ratio (r) = 2

Using the formula: an = a × r(n-1)

192 = a × 2(7-1)

192 = a × 2⁶

192 = a × 64

a = 192/64 = 3

Therefore, the first term is 3.

Question 4: Determine which term of the G.P. 2, 6, 18, 54, ... is 1458.

Solution:

Given G.P.: 2, 6, 18, 54, ...

First term (a) = 2

Common ratio (r) = 6/2 = 3

Let the nth term be 1458

Using: an = a × r(n-1)

1458 = 2 × 3(n-1)

729 = 3(n-1)

3⁶ = 3(n-1)

6 = n-1

n = 7

Therefore, 1458 is the 7th term.

Question 5: Find the 10th term of a G.P. whose first term is 1 and common ratio is 3.

Solution:

Given: First term (a) = 1, Common ratio (r) = 3

Using: an = a × r(n-1)

10th term = 1 × 3(10-1)

= 3⁹

= 19,683

Question 6: Find the sum of the first 10 terms of the G.P. 1, 3, 9, 27, ...

Solution:

Given G.P.: 1, 3, 9, 27, ...

First term (a) = 1, Common ratio (r) = 3, Number of terms (n) = 10

Using the formula: Sn = a(rn - 1)/(r - 1) when r ≠ 1

S₁₀ = 1(3¹⁰ - 1)/(3 - 1)

= (59049 - 1)/2

= 59048/2

= 29,524

Question 7: The sum of the first 5 terms of a G.P. is 242 and the first term is 2. Find the common ratio.

Solution:

Given: First term (a) = 2, Sum of first 5 terms (S₅) = 242

Using: S₅ = a(r⁵ - 1)/(r - 1)

242 = 2(r⁵ - 1)/(r - 1)

121 = (r⁵ - 1)/(r - 1)

121(r - 1) = r⁵ - 1

121r - 121 = r⁵ - 1

r⁵ - 121r + 120 = 0

Testing r = 3:

3⁵ - 121(3) + 120 = 243 - 363 + 120 = 0 ✓

Therefore, the common ratio is 3.

Question 8: Calculate the sum of the first 6 terms of the G.P. whose first term is 4 and common ratio is 0.5.

Solution:

Given: a = 4, r = 0.5, n = 6

Since r < 1, using: Sn = a(1 - rn)/(1 - r)

S₆ = 4(1 - 0.5⁶)/(1 - 0.5)

= 4(1 - 1/64)/(0.5)

= 4(63/64)/0.5

= 4 × 63/64 × 2

= 504/64 = 7.875

Question 9: Find the sum of the first n terms of a G.P. with first term a and common ratio r (r ≠ 1).

Solution:

Let Sn = a + ar + ar² + ar³ + ... + ar(n-1)

Multiplying by r: rSn = ar + ar² + ar³ + ... + arn

Subtracting: Sn - rSn = a - arn

Sn(1 - r) = a(1 - rn)

Therefore: Sn = a(1 - rn)/(1 - r) when r ≠ 1

Question 10: If the sum of the first 4 terms of a G.P. is 30 and the common ratio is 2, find the first term.

Solution:

Given: S₄ = 30, r = 2

Using: S₄ = a(r⁴ - 1)/(r - 1)

30 = a(2⁴ - 1)/(2 - 1)

30 = a(16 - 1)/1

30 = 15a

a = 2

Question 11: Insert three geometric means between 3 and 243.

Solution:

Let the G.P. be: 3, G₁, G₂, G₃, 243

This forms a 5-term G.P. with first term a = 3 and 5th term = 243

Using: a₅ = ar⁴

243 = 3 × r⁴

r⁴ = 81 = 3⁴

r = 3

The three geometric means are:

• G₁ = 3 × 3 = 9
• G₂ = 9 × 3 = 27
• G₃ = 27 × 3 = 81

Question 12: If G is the geometric mean between a and b, prove that G = √(ab).

Solution:

If a, G, b are in G.P., then the ratio between consecutive terms is constant.

Therefore: G/a = b/G

Cross multiplying: G² = ab

Taking square root: G = √(ab)

Hence proved.

Question 13: Insert four geometric means between 1 and 81.

Solution:

Let the G.P. be: 1, G₁, G₂, G₃, G₄, 81

This forms a 6-term G.P. with first term a = 1 and 6th term = 81

Using: a₆ = ar⁵

81 = 1 × r⁵

r⁵ = 81 = 3⁴

r = 3(4/5)

The four geometric means are:

• G₁ = 1 × 3(4/5) = 3(4/5)
• G₂ = 1 × 3(8/5) = 3(8/5)
• G₃ = 1 × 3(12/5) = 3(12/5)
• G₄ = 1 × 3(16/5) = 3(16/5)

Question 14: Three numbers are in G.P. and their sum is 21 and product is 216. Find the numbers.

Solution:

Let the three numbers be a/r, a, ar

Given: Sum = a/r + a + ar = 21

Product = (a/r) × a × ar = a³ = 216

From product: a = ∛216 = 6

Substituting in sum: 6/r + 6 + 6r = 21

Multiplying by r: 6 + 6r + 6r² = 21r

6r² - 15r + 6 = 0

2r² - 5r + 2 = 0

(2r - 1)(r - 2) = 0

r = 1/2 or r = 2

If r = 2: numbers are 3, 6, 12

If r = 1/2: numbers are 12, 6, 3

Both represent the same set: 3, 6, 12

Question 15: Insert two geometric means between 5 and 80.

Solution:

Let the G.P. be: 5, G₁, G₂, 80

This forms a 4-term G.P. with first term a = 5 and 4th term = 80

Using: a₄ = ar³

80 = 5 × r³

r³ = 16

r = ∛16 = 2∛2

The two geometric means are:

• G₁ = 5 × 2∛2 = 10∛2
• G₂ = 5 × (2∛2)² = 5 × 4∛4 = 20∛4

Question 16: Find the sum to infinity of the G.P. 8, 4, 2, 1, ...

Solution:

Given G.P.: 8, 4, 2, 1, ...

First term (a) = 8

Common ratio (r) = 4/8 = 1/2

Since |r| = 1/2 < 1, the series converges.

Sum to infinity = a/(1-r) = 8/(1-1/2) = 8/(1/2) = 16

Question 17: If the sum to infinity of a G.P. is 15 and the first term is 10, find the common ratio.

Solution:

Given: a = 10, S∞ = 15

Using: S∞ = a/(1-r)

15 = 10/(1-r)

15(1-r) = 10

15 - 15r = 10

15r = 5

r = 1/3

Question 18: Determine whether the infinite series 1, 1/2, 1/4, 1/8, ... converges or diverges. If it converges, find its sum.

Solution:

Given series: 1, 1/2, 1/4, 1/8, ...

First term (a) = 1

Common ratio (r) = (1/2)/1 = 1/2

Since |r| = 1/2 < 1, the series converges.

Sum to infinity = a/(1-r) = 1/(1-1/2) = 1/(1/2) = 2

Question 19: Find the sum to infinity of the G.P. whose first term is 6 and common ratio is 1/3.

Solution:

Given: a = 6, r = 1/3

Since |r| = 1/3 < 1, the series converges.

Sum to infinity = a/(1-r) = 6/(1-1/3) = 6/(2/3) = 6 × 3/2 = 9

Question 20:  If the sum to infinity of a G.P. is 9 and the first term is 4, find the common ratio.

Solution:

Given: a = 4, S∞ = 9

Using: S∞ = a/(1-r)

9 = 4/(1-r)

9(1-r) = 4

9 - 9r = 4

9r = 5

r = 5/9