Question

If $S$ is the sum, $P$ the product and $R$ the sum of reciprocals of n terms of a G.P., Then prove that $P^2={\left(\frac{S}{R}\right)}^n$

Answer 1

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

$P=a^n\times r^{1+2+3+....+(n-1)}=a^n{r}^{\frac{n(n-1)}{2}}$ since sum of natural numbers$=\frac{n(n+1)}{2}$

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

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

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

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

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

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

$\Rightarrow P^2{R}^n=S^n$ since $S,P,R$ are in G.P(given)

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

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

Hence proved.