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