If the sum of an infinite decreasing G.P. is 3 and sum of the cubes of its terms is 108/13, then common ratio is given by

$\frac{a}{1-r}=3,\frac{a^3}{1-r^3}=\frac{108}{13}$ where $\left|r\right|<1$

$\therefore \frac{27(1-r{)}^3}{1-r^3}=\frac{108}{13}$

$\begin{array}{l}\Rightarrow \frac{1-2r+r^2}{1+r+r^2}=\frac{4}{13}\\ \Rightarrow 13-26r+13r^2=4+4r+4r^2\\ \Rightarrow 9r^2-30r+9=0\Rightarrow 3r^2-10r+3=0.\\ \Rightarrow (3r-1)(r-3)=0\Rightarrow r=1/3[\because |r|<1]\end{array}$