Sequences And Series Ques 80
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) $\frac{2}{3}$
(c) $\frac{2}{9}$
(d) $\frac{1}{3}$
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Answer:
Correct Answer: 80.(b)
Solution:
Formula:
- Let the GP be $a, a r, a r^{2}, a r^{3}, \ldots \infty$; where $a>0$ and $0<r<1$.
Then, according the problem, we have
$ \begin{aligned} 3 & =\frac{a}{1-r} \\ \text { and } \quad \frac{27}{19} & =a^{3}+(a r)^{3}+\left(a r^{2}\right)^{3}+\left(a r^{3}\right)^{3}+\ldots \\ \Rightarrow \quad \frac{27}{19} & =\frac{a^{3}}{1-r^{3}} \quad [\because S _{\infty}=\frac{a}{1-r}] \end{aligned} $
$ \begin{array}{ll} \Rightarrow & \frac{27}{19}=\frac{(3(1-r))^{3}}{1-r^{3}} \quad \quad [\because 3=\frac{a}{1-r} \Rightarrow a=3(1-r) ]\\ \Rightarrow & \frac{27}{19}=\frac{27(1-r)\left(1+r^{2}-2 r\right)}{(1-r)\left(1+r+r^{2}\right)} \end{array} $
$ \left[\because(1-r)^{3}=(1-r)(1-r)^{2}\right] $
$ \begin{aligned} & \Rightarrow \quad r^{2}+r+1=19\left(r^{2}-2 r+1\right) \\ & \Rightarrow \quad 18 r^{2}-39 r+18=0 \\ & \Rightarrow \quad 6 r^{2}-13 r+6=0 \\ & \Rightarrow \quad(3 r-2)(2 r-3)=0 \end{aligned} $
$ \therefore \quad r=\frac{2}{3} \text { or } r=\frac{3}{2} \text { (reject) } \quad[\because 0<r<1] $
BETA
