The sum of first four terms of a geometric progression (G. P.) is ${{65} \over {12}}$ and the sum of their respective reciprocals is ${{65} \over {18}}$. If the product of first three terms of the G.P. is 1, and the third term is $\alpha$, then 2$\alpha$ is _________.

Let the terms are $a,ar,a{r^2},a{r^3}$<br><br>$a + ar + a{r^2} + a{r^3} = {{65} \over {12}}$ ..........(1)<br><br>$${1 \over a} + {1 \over {ar}} + {1 \over {a{r^2}}} + {1 \over {a{r^3}}} = {{65} \over {18}}$$<br><br>$${1 \over a}\left( {{{{r^3} + {r^2} + r + 1} \over {{r^3}}}} \right) = {{65} \over {18}}$$ ...............(2)<br><br>Doing ${{(1)} \over {(2)}},$ <br><br>${a^2}{r^3} = {{18} \over {12}} = {3 \over 2}$<br><br>Also given, $${a^3}{r^3} = 1 \Rightarrow a\left( {{3 \over 2}} \right) = 1 \Rightarrow a = {2 \over 3}$$<br><br>$${4 \over 9}{r^3} = {3 \over 2} \Rightarrow {r^3} = {{{3^3}} \over {{2^3}}} \Rightarrow r = {3 \over 2}$$<br><br>$\alpha = a{r^2} = {2 \over 3}.{\left( {{3 \over 2}} \right)^2} = {3 \over 2}$<br><br>$2\alpha = 3$