**Question:**

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

**Solution:**

Let number are $a, a r, a r^{2}, a r^{3}$

$a \frac{\left(r^{4}-1\right)}{r-1}=\frac{65}{12}$ .............(1)

$\frac{1}{a} \frac{\left(\frac{1}{r^{4}}-1\right)}{\frac{1}{r}-1}=\frac{65}{18}$

$\frac{1}{a r^{3}}\left(\frac{1-r^{3}}{1-r}\right)=\frac{65}{18}$ ..............(2)

$\frac{(1)}{(2)} \Rightarrow a^{2} r^{3}=\frac{3}{2}$

and $\quad a^{3} \cdot r^{3}=1$

$a r=1$

$(\mathrm{ar})^{2} \cdot \mathrm{r}=\frac{3}{2}$

$\mathrm{r}=\frac{3}{2}, \mathrm{a}=\frac{2}{3}$

So, third term $=a r^{2}=\frac{2}{3} \times \frac{9}{4}$

$\alpha=\frac{3}{2}$

$2 \alpha=3$