If the sum of an infinite GP a, ar, ar $^{2}$, ar $^{3}, \ldots$ is 15 and the sum of the squares of its each term is 150 , then the sum of $a r^{2}, a r^{4}, a r^{6}, \ldots$ is :
Correct Option: , 2
Sum of infinite terms :
$\frac{a}{1-r}=15$ .....(i)
Series formed by square of terms:
$a^{2}, a^{2} r^{2}, a^{2} r^{4}, a^{2} r^{6} \ldots . .$
Sum $=\frac{\mathrm{a}^{2}}{1-\mathrm{r}^{2}}=150$
$\Rightarrow \frac{\mathrm{a}}{1-\mathrm{r}} \cdot \frac{\mathrm{a}}{1+\mathrm{r}}=150 \Rightarrow 15 \cdot \frac{\mathrm{a}}{1+\mathrm{r}}=150$
$\Rightarrow \frac{\mathrm{a}}{1+\mathrm{r}}=10$ ......(ii)
by (i) and (ii) $\mathrm{a}=12 ; \mathrm{r}=\frac{1}{5}$
Now series : $a r^{2}, a r^{4}, a r^{6}$
Sum $=\frac{a r^{2}}{1-r^{2}}=\frac{12 \cdot(1 / 25)}{1-1 / 25}=\frac{1}{2}$