Question

Let $S$ be the sum, $P$ be the product and $R$ be the sum of reciprocals of $n$ terms in a G.P. . Then,

• A. $S^2 R^n=P^n$
• B. $R^2 P^n=S^n$
• C. $R^2 S^n=P^n$
• D. $P^2 R^n=S^n$

Answer 1

Answer: (D)

Let the $n$ terms of G.P. is $a, a r, a r^2, a r^3 \ldots, a r^{n-1}$.
Given, $S=$ Sum of $n$ terms
$=a+a r+a r^2+a r^3+\ldots+a r^{n-1}$
$=\frac{a\left(r^n-1\right)}{r-1}......$(i)
$R=$ Sum of the reciprocals of $n$ terms
$=\frac{1}{a}+\frac{1}{a r}+\frac{1}{a r^2}+\ldots+\frac{1}{a r^{n-1}} $
$ =\frac{\frac{1}{a}\left[1-\left(\frac{1}{r}\right)^n\right]}{1-\frac{1}{r}}=\frac{1}{a}\left[\frac{1}{1}-\frac{1}{r^n}\right] \times \frac{1}{\frac{r-1}{r}}$
$ =\frac{1}{a}\left[\frac{r^n-1}{r^n}\right] \times \frac{r}{r-1}$
$\Rightarrow R=\frac{\left(r^n-1\right) r}{a r^n(r-1)} ....$(ii)
and $P=$ Product of $n$ terms
$=a \times a r \times a r^2 \times a r^3 \times \ldots \times a r^{n-1}$
$= a^{1+1+1+\ldots \text { upto } n \text { terms }} r^{1+2+3+\ldots+(n-1) \text { terms }} $
$= a^n r \frac{n(n-1)}{2} $
$ {\left[\because \Sigma n=\frac{n(n+1)}{2}\right] }$
$\Rightarrow P^2=a^{2 n} r^{n(n-1)}.....$(iii)
Now, consider $P^2 R^n=a^{2 n} r^{n(n-1)}\left[\frac{r\left(r^n-1\right)}{a r^n(r-1)}\right]^n$
[using Eqs. (ii) and (iii)]
$=a^{2 n} r^{n(n-1)} \frac{r^n\left(r^n-1\right)^n}{a^n\left(r^n\right)^n(r-1)^n}$
$ =\frac{a^n r^{n^2}\left(r^n-1\right)^n}{r^{n^2}(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=S^n$ [using Eq. (i)]