If $S_{1}, S_{2}, S_{3}$ are the sum of n,2n and 3n terms of the G.P. then Prove that $S_{1}^{2} + S_{2}^{2} = S_{1} (S_{2} + S_{3}) .$

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Solution

$a, a r, {a r}^{2}, {a r}^{3}, . . . . . .$
$S_{1} = S_{n} = \frac{a (r^{n} - 1)}{r - 1}$
$S_{2} = S_{2 n} = \frac{a (r^{2 n} - 1)}{r - 1}$
$S_{3} = S_{3 n} = \frac{a (r^{3 n} - 1)}{r - 1}$
$S_{1}^{2} = \frac{a^{2}}{{(r - 1)}^{2}} {(r^{n} - 1)}^{2}$
$S_{2}^{2} = \frac{a^{2}}{{(r - 1)}^{2}} {(r^{2 n} - 1)}^{2}$
$L . H . S = S_{1}^{2} + S_{2}^{2}$
$= \frac{a^{2}}{{(r - 1)}^{2}} [{(r^{n} - 1)}^{2} + {(r^{2 n} - 1)}^{2}]$
$= \frac{a^{2}}{{(r - 1)}^{2}} [r^{2 n} - {2 r}^{n} + 1 + r^{4 n} - {2 r}^{2 n} + 1]$
$= \frac{a^{2}}{{(r - 1)}^{2}} [r^{4 n} - r^{2 n} - {2 r}^{n} + 2]$
$R . H . S = S_{1} (S_{2} + S_{3})$
$= \frac{a}{r - 1} (r^{n} - 1) [\frac{a}{r - 1} (r^{2 n} - 1) + \frac{a}{r - 1} (r^{3 n} - 1)]$
$= \frac{a^{2}}{{(r - 1)}^{2}} (r^{n} - 1) [r^{2 n} + r^{3 n} - 2]$
$= \frac{a^{2}}{{(r - 1)}^{2}} (r^{3 n} + r^{4 n} - {2 r}^{n} - r^{2 n} - r^{3 n} + 2)$
$= \frac{a^{2}}{{(r - 1)}^{2}} (r^{4 n} - r^{2 n} - {2 r}^{n} + 2)$
$∴$ $L . H . S = R . H . S$
$\Rightarrow S_{1}^{2} + S_{2}^{2} = S_{1} (S_{2} + S_{3})$ (Proved)

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