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If the sum of an infinite $GP$ $a, ar, ar^{2}, a r^{3}, \ldots$ is $15$ and the sum of the squares of its each term is $150 ,$ then the sum of $\mathrm{ar}^{2}, \mathrm{ar}^{4}, \mathrm{ar}^{6}, \ldots$ is :
$\frac{5}{2}$
$\frac{1}{2}$
$\frac{25}{2}$
$\frac{9}{2}$
Solution
Sum of infinite terms :
$\frac{\mathrm{a}}{1-\mathrm{r}}=15….(i)$
Series formed by square of terms:
$\mathrm{a}^{2}, \mathrm{a}^{2} \mathrm{r}^{2}, \mathrm{a}^{2} \mathrm{r}^{4}, \mathrm{a}^{2} r^{6} \ldots \ldots$
$\text { Sum }=\frac{\mathrm{a}^{2}}{1-\mathrm{r}^{2}}=150$
$\Rightarrow \frac{\mathrm{a}}{1-\mathrm{r}} \cdot \frac{\mathrm{a}}{1+\mathrm{r}}=150 \Rightarrow 15 \cdot \frac{\mathrm{a}}{1+\mathrm{r}}=150$
$\Rightarrow \frac{\mathrm{a}}{1+\mathrm{r}}=10 \ldots \ldots \ldots \text { (ii) }$
by $(i)$ and $(ii)$ $\mathrm{a}=12 ; \mathrm{r}=\frac{1}{5}$
Now series : $\mathrm{ar}^{2}, \mathrm{ar}^{4}, \mathrm{ar}^{6}$
$\text { Sum }=\frac{a r^{2}}{1-r^{2}}=\frac{12 \cdot(1 / 25)}{1-1 / 25}=\frac{1}{2}$
