Question

A G.P. consists of an even number of terms. If the sum of all the terms is five times the sum of the terms occupying odd places, the common ratio is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is C $4$

Let total no. of terms of GP be $2n$

$\because S_{2n}=5$ (sum of all terms occupying odd places)

$\Rightarrow (a+ar+a{r}^2+......a{r}^{2n-1})=5(a+a{r}^2+a{r}^4+....a{r}^{2n-2})$

where $a$ is first term of GP and $r$ is common ratio.

$\Rightarrow \frac{a(r^{2n}-1)}{r-1}=\frac{5a(r^{2n}-1)}{r^2-1}$

$\Rightarrow \frac{(r^{2n}-1)}{r-1}-\frac{5(r^{2n}-1)}{(r+1)(r-1)}=0$

$\Rightarrow (r^{2n}-1)\left(\frac{r+1-5}{(r-1)(r+1)}\right)=0$

$\Rightarrow \frac{(r^{2n}-1)}{(r-1)(r+1)}(r-4)=0$

$\because r\ne \pm 1$

$\therefore r=4$

$\therefore$ Common ratio of the GP $=4$