Mathematics - Sequence & Series Question with Solution | TestHub

Sum of infinite terms of series a $+ar+a{r}^2+$ $=15$
$\therefore \frac{a}{1-r}=15$ ...(i)
Sequence formed by square of terms: $a^2,a^2{r}^2,a^2{r}^4,a^2{r}^6\dots$
$a^2+a^2{r}^2+\cdots =150$
Sum $=\frac{a^2}{1-r^2}=150$
$\Rightarrow \frac{a}{1-r}\cdot \frac{a}{1+r}=15\Rightarrow 15\cdot \frac{a}{1+r}=150$ $\left(\because \frac{\mathrm{a}}{1-\mathrm{r}}=15\right)$
$\Rightarrow \frac{a}{1+r}=10$ ...(ii)
On dividing equation (i) by (ii)
$\frac{1+\mathrm{r}}{1-\mathrm{r}}=\frac{15}{10}$
or $r=\frac{1}{5}\Rightarrow a=12$
Now series : $a{r}^2+a{r}^4+a{r}^6+\dots$
Sum $=\frac{a^2}{1-r^2}=\frac{12\cdot \left(\frac{1}{25}\right)}{1-\frac{1}{25}}=\frac{1}{2}$