Question

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

Answer 1

Let G.P. be $a,ar,a{r}^2,....a{r}^{n-1}$ (n terms)

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

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

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

Product$=P=a.ar.a{r}^2....a{r}^{n-1}$

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

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

Sum of reciprocals$=R=\frac{1}{a}+\frac{1}{ar}+....+\frac{1}{a{r}^{n-1}}$

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

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

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

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

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

$=S^n$

$\therefore P^2{R}^n=S^n$