# 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 ________ Option: 1 2 Option: 2 -2 Option: 3 -3 Option: 4 3

\begin{aligned} &\text {Let number are a, ar, ar }^{2} \text { , ar }^{3}\\ &\mathrm{a} \frac{\left(\mathrm{r}^{4}-1\right)}{\mathrm{r}-1}=\frac{65}{12} \qquad\ldots(1) \end{aligned}

$\\\frac{1}{\mathrm{a}} \frac{\left(\frac{1}{\mathrm{r}^{4}}-1\right)}{\frac{1}{\mathrm{r}}-1}=\frac{65}{18} \\ \\\frac{1}{\mathrm{ar}^{3}}\left(\frac{1-\mathrm{r}^{3}}{1-\mathrm{r}}\right)=\frac{65}{18}\qquad\ldots(2)$

$\\\frac{(1)}{(2)} \Rightarrow a^{2} r^{3}=\frac{3}{2} \\\\ \text { and } \quad a^{3} \cdot r^{3}=1 \\\\ a r=1 \\\\ (a r)^{2} \cdot r=\frac{3}{2} \\ \\r=\frac{3}{2}, a=\frac{2}{3}$

\begin{aligned} \text {So, third term }=a r^{2} &=\frac{2}{3} \times \frac{9}{4} \\ \alpha &=\frac{3}{2} \\ 2 \alpha &=3 \end{aligned}

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