Let S be the sum,
Question:

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

Solution:

Let the G.P. be $a, a r, a r^{2}, a r^{3}, \ldots a r^{n-1} \ldots$

According to the given information,

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

$\mathrm{P}=a^{n} \times v^{1+2+\ldots+n-1}$

$=a^{n} r^{\frac{a(n-1)}{2}}$ $\left[\because\right.$ Sum of first $n$ natural numbers is $\left.n \frac{(n+1)}{2}\right]$

$\mathrm{R}=\frac{1}{a}+\frac{1}{a r}+\ldots+\frac{1}{a r^{n-1}}$

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

$=\frac{1\left(r^{n}-1\right)}{(r-1)} \times \frac{1}{a r^{n-1}} \quad\left[\because 1, r, \ldots r^{n-1}\right.$ forms a G.P $]$

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

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

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

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

$=\mathrm{S}^{n}$

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

 

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