If $S_{1}, S_{2}$ and S denote respectively the sum of n, $2 n$ and infinite number of terms of a G.P. then prove that $S_{1} (S_{1} - S) = (S_{1} - S_{2}) S$ .

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Solution

$S_{1} = \frac{a (1 - r^{n})}{1 - r}$
$S_{2} = \frac{a (1 - r^{2 n})}{1 - r}$
$S = \frac{a}{1 - r}$
$S_{1} - S_{2} = \frac{a (r^{2 n} - r^{n})}{r - 1}$
$⟹ \frac{a r^{n} (r^{n} - 1)}{1 - r}$
$S (S_{1} - S_{2}) = \frac{a^{2} r^{n} (r^{n} - 1)}{(1 - r)^{2}}$
$S (S_{1} - S_{2}) = \frac{S_{1} (- a r^{n})}{1 - r}$
$S (S_{1} - S_{2}) = \frac{S_{1} a (1 - r^{n} - 1)}{1 - r}$
$S (S_{1} - S_{2}) = S_{1} (\frac{a (1 - r^{n})}{1 - r} - \frac{a}{1 - r})$
$S (S_{1} - S_{2}) = \frac{S_{1} a (1 - r^{n} - 1)}{1 - r}$
$S (S_{1} - S_{2}) = S_{1} (S_{1} - S)$

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If S1,S2 and S denote respectively the sum of n, 2n and infinite number of terms of a G.P. then prove that S1(S1−S)=(S1−S2)S.

S1=a(1−rn)1−rS2=a(1−r2n)1−rS=a1−rS1−S2=a(r2n−rn)r−1⟹arn(rn−1)1−rS(S1−S2)=a2rn(rn−1)(1−r)2S(S1−S2)=S1(−arn)1−rS(S1−S2)=S1a(1−rn−1)1−rS(S1−S2)=S1(a(1−rn)1−r−a1−r)S(S1−S2)=S1a(1−rn−1)1−rS(S1−S2)=S1(S1−S)

If $S_{1}, S_{2}$ and S denote respectively the sum of n, $2 n$ and infinite number of terms of a G.P. then prove that $S_{1} (S_{1} - S) = (S_{1} - S_{2}) S$ .