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