# Solve the following

Question:

If $S_{1}, S_{2}, S_{3}$ be respectively the sums of $n, 2 n, 3 n$ terms of a G.P., then prove that $S_{1}^{2}+S_{2}^{2}=S_{1}\left(S_{2}+S_{3}\right)$.

Solution:

Let a be the first term and r be the common ratio of the given G.P.

Sum of $n$ terms, $S_{1}=a\left(\frac{r^{n}-1}{r-1}\right)$           ...(1)

Sum of $2 n$ terms, $S_{2}=a\left(\frac{r^{2 n}-1}{r-1}\right)$

$\Rightarrow S_{2}=a\left[\frac{\left(r^{n}\right)^{2}-1^{2}}{r-1}\right]$

$\Rightarrow S_{2}=a\left[\frac{\left(r^{n}-1\right)\left(r^{n}+1\right)}{r-1}\right]$

$\Rightarrow S_{2}=S_{1}\left(r^{n}+1\right)$          ...(2)

And, $s$ um of $3 n$ terms, $S_{3}=a\left(\frac{r^{3 n}-1}{r-1}\right)$

$\Rightarrow S_{3}=a\left[\frac{\left(r^{n}\right)^{3}-1^{3}}{r-1}\right]$

$\Rightarrow S_{3}=a\left[\frac{\left(r^{n}-1\right)\left(r^{2 n}+r^{n}+1\right)}{r-1}\right]$

$\Rightarrow S_{3}=S_{1}\left(r^{2 n}+r^{n}+1\right)$    ....(3)

Now, LHS $=\left(S_{1}\right)^{2}+\left(S_{2}\right)^{2}$

$=\left(S_{1}\right)^{2}+\left[S_{1}\left(r^{n}+1\right)\right]^{2} \quad[$ Using $(2)]$

$=\left(S_{1}\right)^{2}\left[1+\left(r^{n}+1\right)^{2}\right]$

$=\left(S_{1}\right)^{2}\left[1+r^{2 n}+2 r^{n}+1\right]$

$=\left(S_{1}\right)^{2}\left[r^{2 n}+r^{n}+1+r^{n}+1\right]$

$=\left(S_{1}\right)\left[S_{1}\left(r^{2 n}+r^{n}+1\right)+S_{1}\left(r^{n}+1\right)\right]$

$=\left(S_{1}\right)\left[S_{2}+S_{3}\right]$

= RHS

Hence proved.