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 'a' be the first term and 'r' be the common ratio of G.P.

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

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

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

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

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

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

$=\frac{\frac{1}{a}[1-{\left(\frac{1}{r}\right)}^n]}{1-\frac{1}{r}}=\frac{r}{a(r-1)\left[\frac{r^n-1}{r^n}\right]}$

Now,

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

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

$=\frac{a^n(r^n-1{)}^n}{(r-1{)}^n}={\left[\frac{a(r^n-1)}{r-1}\right]}^n=S^n$

Thus, $P^2{R}^n=S^n$