Question

If the sum of an infinite GP $a,ar,a{r}^2,a{r}^3,...$ is $15$ and the sum of the squares of its each term is $150$, then the sum of $a{r}^2,a{r}^4,a{r}^6,...$ is :

• A. $\frac{5}{2}$
• B. $\frac{1}{2}$
• C. $\frac{25}{2}$
• D. $\frac{9}{2}$

Answer 1

Answer: (B)

Sum of infinite terms:
$\frac{a}{1-r}=15.......(i)$
Series formed by square of terms:
$a^2,a^2{r}^2,a^2{r}^4,a^2{r}^6\dots$
Sum $=\frac{a^2}{1-r^2}=150$
$\Rightarrow \frac{a}{1-r}\cdot \frac{a}{1+r}=150$
$\Rightarrow 15\cdot \frac{a}{1+r}=150$
$\Rightarrow \frac{a}{1+r}=10\dots \dots \dots$ (ii)
by (i) and (ii) $a=12;r=\frac{1}{5}$
Now series : $a{r}^2,a{r}^4,a{r}^6$
Sum $=\frac{a{r}^2}{1-r^2}=\frac{12\cdot (1/25)}{1-1/25}=\frac{1}{2}$