Mathematics - Sequence & Series Question with Solution | TestHub

Given:

$a+ar+a{r}^2+a{r}^3+......\infty =15$

$\Rightarrow \frac{a}{1-r}=15 ....\left(1\right)$ 

And,

$a^2+a^2{r}^2+a^2{r}^4+a^2{r}^6+......\infty =150$

$\Rightarrow \frac{a^2}{1-r^2}=150$

$\Rightarrow \left(\frac{a}{1+r}\right)\left(\frac{a}{1-r}\right)=150$

$\Rightarrow \frac{a}{(1+r)}=10 ...\left(2\right)$

Solving equation $\left(1\right)$ and $\left(2\right)$, we get

$r=\frac{1}{5}\Rightarrow a=12$

Now,

$a{r}^2+a{r}^4+a{r}^6......\infty$

$=\frac{a{r}^2}{1-r^2}=\frac{12\times {\displaystyle \frac{1}{25}}}{1-{\displaystyle \frac{1}{25}}}=\frac{1}{2}$