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

Answer 1

Let the first four terms be $a,ar,a{r}^2,a{r}^3.$
$a+ar+a{r}^2+a{r}^3=\frac{65}{12}\cdots \left(1\right)$
and $\frac{1}{a}+\frac{1}{ar}+\frac{1}{a{r}^2}+\frac{1}{a{r}^3}=\frac{65}{18}$
$\Rightarrow \frac{1}{a}\left(\frac{r^3+r^2+r+1}{r^3}\right)=\frac{65}{18}\cdots \left(2\right)$
$\frac{\left(1\right)}{\left(2\right)},$ we get
$a^2{r}^3=\frac{18}{12}=\frac{3}{2}$

Also, $a^3{r}^3=1$
$\Rightarrow a\left(\frac{3}{2}\right)=1\Rightarrow a=\frac{2}{3}$
$\frac{4}{9}{r}^3=\frac{3}{2}$
$\Rightarrow r^3=\frac{3^3}{2^3}\Rightarrow r=\frac{2}{3}$
$\alpha =a{r}^2=\frac{2}{3}.{\left(\frac{3}{2}\right)}^2=\frac{3}{2}$
$\therefore 2\alpha =3$