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. $\frac{2}{3}$
• D. $\frac{1}{3}$

Answer 1

Answer: (C)

$\frac{2}{3}$

$\Rightarrow \frac{a}{1-r}=3$ .....(1)

$\therefore a^3=27{(1-r)}^3$

$\frac{a^3}{1-r^3}=\frac{27}{19}\Rightarrow \frac{27(1-r{)}^3}{1-r^3}=\frac{27}{19}$

$\Rightarrow 6r^2-13r+6=0$

$\Rightarrow r=\frac{2}{3}$ as $|r|<1$