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14. Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that $P^2 R^n = S ^n$

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Ler there be a GP $=a,ar,ar^2,ar^3,....................$

According to given information,

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

$P=a^n \times r^{(1+2+...................n-1)}$

$P=a^n \times r^{\frac{n(n-1)}{2}}$

$R=\frac{1}{a}+\frac{1}{ar}+\frac{1}{ar^2}+..................\frac{1}{ar^{n-1}}$

$R=\frac{r^{n-1}+r^{n-2}+r^{n-3}+..............r+1}{a.r^{n-1}}$

$R=\frac{1}{a.r^{n-1}}\times \frac{1(r^n-1)}{r-1}$

$R= \frac{(r^n-1)}{a.r^{n-1}.(r-1)}$

To prove : $P^2 R^n = S ^n$

LHS : $P^2 R^n$

$= a^{2n}.r^{n(n-1)}\frac{(r^n-1)^n}{a^n.r^{n(n-1)}.(r-1)^n}$

$= a^{n} \frac{(r^n-1)^n}{(r-1)^n}$

$= \left ( \frac{a(r^n-1)}{(r-1)} \right )^{n}$

$=S^n=RHS$

Hence proved

Posted by

seema garhwal

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