Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# The Sum of an Infinite G.P. is 4 and the Sum of the Cubes of Its Terms is 92. the Common Ratio of the Original G.P. is - Mathematics

MCQ

The sum of an infinite G.P. is 4 and the sum of the cubes of its terms is 92. The common ratio of the original G.P. is

• (a) 1/2

• (b) 2/3

• (c) 1/3

• (d) −1/2

#### Solution

(a) 1/2

$\text{ Let the G . P . be a, ar }, a r^2 , a r^3 , . . . , \infty .$
$S_\infty = 4$
$\Rightarrow \frac{a}{1 - r} = 4 (i)$
$\text{ Also, sum of the cubes }, S_1 = 92$
$\Rightarrow \frac{a^3}{\left( 1 - r^3 \right)} = 92 (ii)$
$\text{ Putting the value of a from } (i) \text{ to } (ii):$
$\Rightarrow \frac{\left( 4(1 - r) \right)^3}{\left( 1 - r^3 \right)} = 92$
$\Rightarrow \frac{64(1 - r )^3}{\left( 1 - r^3 \right)} = 92$
$\Rightarrow \frac{\left( 1 - r \right)^3}{\left( 1 - r \right)\left( 1 + r + r^2 \right)} = \frac{92}{64}$
$\Rightarrow \frac{\left( 1 - r \right)^2}{\left( 1 + r + r^2 \right)} = \frac{23}{16}$
$\Rightarrow 16\left( 1 - 2r + r^2 \right) = 23\left( 1 + r + r^2 \right)$
$\Rightarrow 7 r^2 + 55r + 7 = 0$
$\text{ Using the quadratic formula }:$
$\Rightarrow r = \frac{- 55 + \sqrt{{55}^2 - 4 \times 7 \times 7}}{2 \times 7}$
$\Rightarrow r = \frac{- 55 + \sqrt{{55}^2 - {14}^2}}{14}$
$\Rightarrow r = \frac{- 55 + \sqrt{2829}}{14}$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 20 Geometric Progression
Q 10 | Page 57
The text contains a mathematical question in the form of a multiple-choice question (MCQ) about the common ratio of an infinite geometric progression (G.P.), along with a detailed solution that leads to the answer.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# The Sum of an Infinite G.P. is 4 and the Sum of the Cubes of Its Terms is 92. the Common Ratio of the Original G.P. is - Mathematics

MCQ

The sum of an infinite G.P. is 4 and the sum of the cubes of its terms is 92. The common ratio of the original G.P. is

• (a) 1/2

• (b) 2/3

• (c) 1/3

• (d) −1/2

#### Solution

(a) 1/2

$\text{ Let the G . P . be a, ar }, a r^2 , a r^3 , . . . , \infty .$
$S_\infty = 4$
$\Rightarrow \frac{a}{1 - r} = 4 (i)$
$\text{ Also, sum of the cubes }, S_1 = 92$
$\Rightarrow \frac{a^3}{\left( 1 - r^3 \right)} = 92 (ii)$
$\text{ Putting the value of a from } (i) \text{ to } (ii):$
$\Rightarrow \frac{\left( 4(1 - r) \right)^3}{\left( 1 - r^3 \right)} = 92$
$\Rightarrow \frac{64(1 - r )^3}{\left( 1 - r^3 \right)} = 92$
$\Rightarrow \frac{\left( 1 - r \right)^3}{\left( 1 - r \right)\left( 1 + r + r^2 \right)} = \frac{92}{64}$
$\Rightarrow \frac{\left( 1 - r \right)^2}{\left( 1 + r + r^2 \right)} = \frac{23}{16}$
$\Rightarrow 16\left( 1 - 2r + r^2 \right) = 23\left( 1 + r + r^2 \right)$
$\Rightarrow 7 r^2 + 55r + 7 = 0$
$\text{ Using the quadratic formula }:$
$\Rightarrow r = \frac{- 55 + \sqrt{{55}^2 - 4 \times 7 \times 7}}{2 \times 7}$
$\Rightarrow r = \frac{- 55 + \sqrt{{55}^2 - {14}^2}}{14}$
$\Rightarrow r = \frac{- 55 + \sqrt{2829}}{14}$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 20 Geometric Progression
Q 10 | Page 57
user
The text contains a mathematical question in the form of a multiple-choice question (MCQ) about the common ratio of an infinite geometric progression (G.P.), along with a detailed solution that leads to the answer.

TYPE: SINGLE_QUESTION_WITH_ANSWER