Thank you for reporting, we will resolve it shortly
Q.
If first term of a decreasing infinite G.P. is 1 and sum is $S$, then sum of squares of its terms is-
Sequences and Series
Solution:
First term of infinite G.P. $=1$ and sum $=5$
Let common ratio is $r$
So $S=\frac{1}{1-r}$
Let sum of squares of its terms is $S ^{\prime}$.
$\therefore S^{\prime}=1+r^2+r^4 \ldots \ldots \infty$
$\therefore S^{\prime}=\frac{1}{1-r^2}=\frac{1}{(1+r)(1-r)}=\frac{S}{\left(1+\frac{S-1}{S}\right)} \left[\because S=\frac{1}{1-r}\right]$
$\therefore S^{\prime}=\frac{S}{\left(\frac{2 S-1}{S}\right)}=\frac{S^2}{2 S-1}$