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

Find the required value

Let the terms of the GP be $a,ar,a{r}^2,a{r}^3$

Therefore from the given condition we have

The sum of the first four terms is

$$\begin{gathered} a+ar+a{r}^2+a{r}^3=\frac{65}{12} \\ \Rightarrow a\left(1+r+r^2+r^3\right)=\frac{65}{12}.........\left(i\right) \end{gathered}$$

The sum of the reciprocal of the first four term is

$$\begin{gathered} \frac{1}{a}+\frac{1}{ar}+\frac{1}{a{r}^2}+\frac{1}{a{r}^3}=\frac{65}{18} \\ \Rightarrow \frac{1}{a}\left(1+\frac{1}{r}+\frac{1}{r^2}+\frac{1}{r^3}\right)=\frac{65}{18} \\ \Rightarrow \frac{1}{a}\left(\frac{r^3+r^2+r+1}{r^3}\right)=\frac{65}{18}.....................\left(ii\right) \end{gathered}$$

Dividing $\left(i\right)$ by $\left(ii\right)$ we get,

$$\begin{gathered} a^2{r}^3=\frac{18}{12} \\ \Rightarrow a^2{r}^3=\frac{3}{2}...........\left(iii\right) \end{gathered}$$

Given that the product of the first three terms is $1$

$$\begin{gathered} \Rightarrow a^3{r}^3=1 \\ \Rightarrow r^3=\frac{1}{a^3}............\left(iv\right) \end{gathered}$$

Using result $\left(iii\right)\&\left(iv\right)$ we get

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

Putting the value of $a$ in $\left(iv\right)$ we get,

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

From given

$$\begin{gathered} \alpha =a{r}^2 \\ =\left(\frac{2}{3}\right){\left(\frac{3}{2}\right)}^2 \\ =\frac{3}{2} \\ \therefore 2\alpha =3 \end{gathered}$$

Hence, the value of $2\alpha$ is $3$.