If $S$ be the sum, $P$ the product $& R$ the sum of the reciprocals of $n$ terms of a GP, find the value of $P^{2} {(\frac{R}{S})}^{n} .$

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Solution

Let

$a, a r, a r^{2}, . . . . . . . . ., a r^{n - 1}$ be the n terms of G.P.

$S = a + a r + a r^{2} + . . . . . . . . . + a r^{n - 1}$

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

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

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

multiply both sides by $a^{2} r^{n - 1}$

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

$= a r^{n - 1} + a r^{n - 2} + . . . . . . . . . . + a r^{2} + a r + a$

$R (a^{2} r^{n - 1}) = a, a r, a r^{2}, . . . . . . . . ., a r^{n - 1}$

$R (a^{2} r^{n - 1}) = S$

$\frac{S}{R} = a^{2} r^{n - 1}$

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

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

$∴ {(\frac{S}{R})}^{n} = P^{2}$

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