Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P2Rn = Sn

Let the terms in G.P. be a, ar, ar², ar³, … ar^{n – 1}…

Form the question, we have

\( s_n=\dfrac{a(1-r^n)}{1-r}\)

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

Since the sum of first n natural numbers is

\( \dfrac{n(n+1)}{2}\)

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

\( R=\dfrac{1}{a}+\dfrac{1}{ar}+..+\dfrac{1}{ar^{n-1}}\)

\( \dfrac{r^{n-1}+r^{n-2}+...r+1}{ar^{n-1}}\)

Since 1,r,.... \( r^{n-1}\) forms a G.P.,

\( R=\dfrac{1(r^n-1)}{(r-1)}\times \dfrac{1}{ar^{n-1}}\)

= \( \dfrac{(r^n-1)}{ar^{n-1}(r-1)}\)

Then,

\( P^2R^2=a^{2n}r^{n(n-1)}\)
\( \dfrac{(r^n-1)^n}{a^nr^{n-1}(r-1)}\)

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

Therefore, \( P^2R^2=S^n\)