SATHEE: Sequences and Series 4 Question 3

Sequences and Series 4 Question 3

3.

The sum of an infinite geometric series with positive terms is $3$ and the sum of the cubes of its terms is $\frac{27}{19}$ . Then, the common ratio of this series is

(2019 Main, 11 Jan I)

(a) $\frac{4}{9}$

(b) $\frac{2}{3}$

(d) $\frac{1}{3}$

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Answer:

Correct Answer: 3. (b)

Solution:

Let the GP be $a, a r, a r^{2}, a r^{3}, \dots \infty$ ; where $a > 0$ and $0 < r < 1$ .

Then, according the problem, we have

$\begin{aligned} 3 & = \frac{a}{1 - r} \\ and \frac{27}{19} & = a^{3} + (a r)^{3} + {(a r^{2})}^{3} + {(a r^{3})}^{3} + \dots \\ \Rightarrow \frac{27}{19} & = \frac{a^{3}}{1 - r^{3}} [∵ S_{\infty} = \frac{a}{1 - r}] \end{aligned}$

$\begin{array}{ll} \Rightarrow & \frac{27}{19} = \frac{(3 (1 - r))^{3}}{1 - r^{3}} [∵ 3 = \frac{a}{1 - r} \Rightarrow a = 3 (1 - r)] \\ \Rightarrow & \frac{27}{19} = \frac{27 (1 - r) (1 + r^{2} - 2 r)}{(1 - r) (1 + r + r^{2})} \end{array}$

$[∵ (1 - r)^{3} = (1 - r) (1 - r)^{2}]$

$\begin{aligned} \Rightarrow r^{2} + r + 1 = 19 (r^{2} - 2 r + 1) \\ \Rightarrow 18 r^{2} - 39 r + 18 = 0 \\ \Rightarrow 6 r^{2} - 13 r + 6 = 0 \\ \Rightarrow (3 r - 2) (2 r - 3) = 0 \end{aligned}$

$∴ r = \frac{2}{3} or r = \frac{3}{2} (reject) [∵ 0 < r < 1]$

Sequences and Series 4 Question 3

3.

Answer:

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Sequences and Series 4 Question 3

3.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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