If S be the sum,P be the product and R the sum of reciprocals of n terms in a G.P,prove that P^2=(frac{S}R)^n?

Mar 16, 2018

See below.

Explanation:

The right formulation is

${P}^{2} = {\left(\frac{S}{R}\right)}^{n + 1}$

because

$S = {\sum}_{k = 0}^{n} {a}^{k} = \frac{{a}^{n + 1} - 1}{a - 1}$

$R = {\sum}_{k = 0}^{n} {a}^{-} k = \frac{\frac{1}{a} ^ \left(n + 1\right) - 1}{\frac{1}{a} - 1} = \frac{a \left(1 - {a}^{n + 1}\right)}{\left(1 - a\right) {a}^{n + 1}}$

then

$\frac{S}{R} = {a}^{n}$

and

$P = {\prod}_{k = 0}^{n} {a}^{k} = {a}^{{\sum}_{k = 0}^{n} k} = {a}^{\frac{n \left(n + 1\right)}{2}}$

hence

${P}^{2} = {a}^{n \left(n + 1\right)} \ne {a}^{n \times n}$

so the exact formulation is

${P}^{2} = {\left(\frac{S}{R}\right)}^{n + 1}$