The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is 27/19. Then the common ratio of this series is

Let $a$ be the first term and $r$ be the common ratio of the G.P.

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

Now, sum of the cubes of its terms is   $\frac{a^3}{1-r^3}=\frac{27}{19}$

$\Rightarrow \frac{27(1-r{)}^3}{1-r^3}=\frac{27}{19}$           $\text{ [From (i)] }$                                   

$\begin{array}{l}\Rightarrow \frac{(1-r)\left(1+r^2-2r\right)}{(1-r)\left(1+r+r^2\right)}=\frac{1}{19}\\ \Rightarrow 19+19r^2-38r=1+r+r^2\Rightarrow 6r^2-13r+6=0\\ \Rightarrow r=\frac{3}{2}\text{ or }\frac{2}{3}\end{array}$

Since, $\left|r\right|<1,$so $r=\frac{2}{3}$.