• # question_answer If the sum of the $n$terms of G.P. is $S$ product is $P$ and sum of their inverse is $R$, than ${{P}^{2}}$ is equal to  [IIT 1966; Roorkee 1981] A) $\frac{R}{S}$ B) $\frac{S}{R}$ C) ${{\left( \frac{R}{S} \right)}^{n}}$ D) ${{\left( \frac{S}{R} \right)}^{n}}$

Correct Answer: D

Solution :

Given that sum $S=\frac{a({{r}^{n}}-1)}{r-1}=\frac{a\,(1-{{r}^{n}})}{1-r}$ ......(i) $P=a(ar)(a{{r}^{2}})..........(a{{r}^{n-1}})={{a}^{n}}{{r}^{1+2+.........+(n-1)}}$ $={{a}^{n}}{{r}^{(n-1)n/2}}\ i.e.,\ {{P}^{2}}={{a}^{2n}}{{r}^{n(n-1)}}$     ......(ii) and $R=\frac{1}{a}+\frac{1}{ar}+\frac{1}{a{{r}^{2}}}+..........$upto $\frac{1}{49}$ terms            $=\frac{1}{a}\left( 1+\frac{1}{r}+\frac{1}{{{r}^{2}}}+.........\text{upto}\ n\ \text{terms} \right)$         $=\frac{\frac{1}{a}\left[ {{\left( \frac{1}{r} \right)}^{n}}-1 \right]}{\left( \frac{1}{r}-1 \right)}\left( \because \ \frac{1}{r}>1 \right)$ if $r<1$        $=\frac{(1-{{r}^{n}})}{a{{r}^{n-1}}(1-r)}$ ....... (iii) Therefore , $\frac{S}{R}=\frac{a(1-{{r}^{n}})}{1-r}\times \frac{a{{r}^{n-1}}(1-r)}{(1-{{r}^{n}})}={{a}^{2}}{{r}^{n-1}}$ or ${{\left( \frac{S}{R} \right)}^{n}}={{({{a}^{2}}{{r}^{n-1}})}^{n}}={{a}^{2n}}{{r}^{n(n-1)}}={{P}^{2}}$.

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