The sum of first four terms of a geometric progression (G.P.) is $\frac{65}{12}$ and the sum of their respective reciprocals is $\frac{65}{18} .$ If the product of first three terms of the G.P. is 1, and the third term is $\alpha$, then $2 \alpha$ is
$a, a r, a r^{2}, a r^{3}$
$a+a r+a r^{2}+a r^{3}=\frac{65}{12} \ldots(1)$
$\frac{1}{a}+\frac{1}{a r}+\frac{1}{a r^{2}}+\frac{1}{a r^{3}}=\frac{65}{18}$
$\frac{1}{a}\left(\frac{r^{3}+r^{2}+r+1}{r^{3}}\right)=\frac{65}{18} \ldots(2)$
$\frac{(i)}{(i i)}, a^{2} r^{3}=\frac{18}{12}=\frac{3}{2}$
$a^{3} r^{3}=1 \Rightarrow a\left(\frac{3}{2}\right)=1 \Rightarrow a=\frac{2}{3}$
$\frac{4}{9} r^{3}=\frac{3}{2} \Rightarrow r^{3}=\frac{3^{3}}{2^{3}} \Rightarrow r=\frac{3}{2}$
$\alpha=a r^{2}=\frac{2}{3} \cdot\left(\frac{3}{2}\right)^{2}=\frac{3}{2}$
$2 \alpha=3$