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Question
If S1, S2, S3 be respectively the sums of n, 2n, 3n terms of a G.P., then prove that \[S_1^2 + S_2^2\] = S1 (S2 + S3).
Solution
Let a be the first term and r be the common ratio of the given G.P.
\[\text {Sum of n terms, } S_1 = a\left( \frac{r^n - 1}{r - 1} \right) . . . \left( 1 \right)\]
\[\text { Sum of 2n terms }, S_2 = a\left( \frac{r^{2n} - 1}{r - 1} \right)\]
\[ \Rightarrow S_2 = a\left[ \frac{\left( r^n \right)^2 - 1^2}{r - 1} \right]\]
\[ \Rightarrow S_2 = a\left[ \frac{\left( r^n - 1 \right)\left( r^n + 1 \right)}{r - 1} \right]\]
\[ \Rightarrow S_2 = S_1 \left( r^n + 1 \right) . . . . \left( 2 \right)\]
\[\text { And, sum of 3n terms }, S_3 = a\left( \frac{r^{3n} - 1}{r - 1} \right)\]
\[ \Rightarrow S_3 = a\left[ \frac{\left( r^n \right)^3 - 1^3}{r - 1} \right]\]
\[ \Rightarrow S_3 = a\left[ \frac{\left( r^n - 1 \right)\left( r^{2n} + r^n + 1 \right)}{r - 1} \right]\]
\[ \Rightarrow S_3 = S_1 \left( r^{2n} + r^n + 1 \right) . . . \left( 3 \right)\]
\[\text { Now, LHS }= \left( S_1 \right)^2 + \left( S_2 \right)^2 \]
\[ = \left( S_1 \right)^2 + \left[ S_1 \left( r^n + 1 \right) \right]^2 \left[ \text { Using } \left( 2 \right) \right]\]
\[ = \left( S_1 \right)^2 \left[ 1 + \left( r^n + 1 \right)^2 \right]\]
\[ = \left( S_1 \right)^2 \left[ 1 + r^{2n} + 2 r^n + 1 \right]\]
\[ = \left( S_1 \right)^2 \left[ r^{2n} + r^n + 1 + r^n + 1 \right]\]
\[ = \left( S_1 \right)\left[ S_1 \left( r^{2n} + r^n + 1 \right) + S_1 \left( r^n + 1 \right) \right]\]
\[ = \left( S_1 \right)\left[ S_2 + S_3 \right] \left[ Using \left( 2 \right) and \left( 3 \right) \right]\]
= RHS
Hence proved .
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