Question

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 :

• A. $\frac{4}{9}$
• B. $\frac{2}{9}$
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

The correct option is D $\frac{2}{3}$

Let $a,ar,a{r}^2,..$ be the G.P.
Given $\frac{a}{1-r}=3$ (Assuming $|r|<1)$
$\Rightarrow {\left(\frac{a}{1-r}\right)}^3=3^3\dots \left(1\right)$

According to the question,
$\frac{a^3}{1-r^3}=\frac{27}{19}\dots \left(2\right)$

Dividing $\left(2\right)$ by $\left(1\right)$, we get
$\frac{(1-r{)}^3}{1-r^3}=\frac{27}{19\times 27}$

$\Rightarrow \frac{(1-r{)}^2}{1+r+r^2}=\frac{1}{19}$

$\Rightarrow 6r^2-13r+6=0$
$\Rightarrow r=\frac{2}{3}(\because |r|<1)$