Question

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

Answer 1

Let the given GP be a, ar, $a{r}^2$, . . ., $a{r}^{(n-1)}$. Then,

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

and $P=[a\times ar\times a{r}^2\times \cdots \times a{r}^{(n-1)}]$

$\Rightarrow P=a^n{r}^{[1+2+3+\cdots +(n-1\left)\right]}=a^n{r}^{\frac{1}{2}(n-1)n}$

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

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

And $R=\{\frac{1}{a}+\frac{1}{ar}+\cdots +\frac{1}{a{r}^{n-1}}\}=\left(\frac{1}{a}\right).\frac{\{\frac{1}{r^n}-1\}}{\{\frac{1}{r}-1\}}$

$\Rightarrow R=\left(\frac{r}{a}\right).\frac{(1-r^n)}{(1-r).r^n}$

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

On dividing (i) by (iii), we get

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

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

[using (ii)].

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