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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 ....... .
$5$
$6$
$2$
$3$
Solution
Let number are $a , ar , ar ^{2}, ar ^{3}$
$a \frac{\left(r^{4}-1\right)}{r-1}=\frac{65}{12}……(1)$
$\frac{1}{a} \frac{\left(\frac{1}{r^{4}}-1\right)}{\frac{1}{r}-1}=\frac{65}{18}$
$\frac{1}{a r^{3}}\left(\frac{1-r^{3}}{1-r}\right)=\frac{65}{18}……(2)$
$\frac{(1)}{(2)} \Rightarrow a ^{2} r ^{3}=\frac{3}{2}$
and $\quad a^{3} \cdot r^{3}=1$
$ar =1$
$(\operatorname{ar})^{2} \cdot r =\frac{3}{2}$
$r=\frac{3}{2}, a=\frac{2}{3}$
So, third term $=\operatorname{ar}^{2}=\frac{2}{3} \times \frac{9}{4}$
$\alpha=\frac{3}{2}$
$2 \alpha=3$
