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

Answer 1

The correct option is D $\frac{1}{2}$

$\because a+ar+a{r}^2+\cdots \infty =15$
$\therefore \frac{a}{1-r}=15\cdots \left(i\right)$
and $a^2+a^2{r}^2+a^2{r}^4+\cdots \infty =150$
$\therefore \frac{a^2}{1-r^2}=150\cdots \left(ii\right)$
Dividing equation $\left(ii\right)$ by square of equation $\left(i\right),$ we get
$\frac{1-r}{1+r}=\frac{2}{3}$
$\Rightarrow 3-3r=2+2r$
$\therefore r=\frac{1}{5}$
Hence, $a=15\times (1-\frac{1}{5})=12$
$\therefore a{r}^2+a{r}^4+a{r}^6+....=\frac{a{r}^2}{1-r^2}=\frac{12\times \frac{1}{25}}{\frac{24}{25}}=\frac{1}{2}$