Question

If $S$ be the sum, $P$ the product and $R$ the sum of the reciprocals of $n$ terms of a G.P, then $P^2$ is equal to:

• A. ${\left(\frac{S}{R}\right)}^n$
• B. $\left(\frac{S}{R}\right)$
• C. ${\left(\frac{R}{S}\right)}^n$
• D. $\left(\frac{R}{S}\right)$

Answer 1

The correct option is A

${\left(\frac{S}{R}\right)}^n$

Explanation for the correct option.

Step 1: Find product of terms

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

We know that sum of n terms in GP i.e; $\mathrm{S}=\mathrm{a}+\mathrm{ar}+a{r}^2+a{r}^3+\dots +a{r}^{\mathrm{n}-1}$ is given by:

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

Then, $\mathrm{P}=\mathrm{a}·{\mathrm{ar}}^{\mathrm{ar}}·{\mathrm{ar}}^2..{\mathrm{ar}}^{\mathrm{n}-1}$

$$\begin{gathered} =a^n{r}^{1+2+3+.n-1} \\ =a^{n_1}{r}^{(n-1)n2} \end{gathered}$$

$\Rightarrow {\mathrm{P}}^2={\mathrm{a}}^{2\mathrm{n}}{\mathrm{r}}^{\mathrm{n}(\mathrm{n}-1)}\dots \left(2\right)$

Step 2: Find the sum of reciprocal

$\begin{array}{rcl}\mathrm{R}& =& \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{ar}}+\frac{1}{{\mathrm{ar}}^2}+\dots ..+\frac{1}{{\mathrm{ar}}^{\mathrm{n}-1}}\\ R& =& \frac{1}{a}·\frac{\left(1-\frac{1}{r^n}\right)}{\left(1-\frac{1}{r}\right)}\\ & =& \frac{\left(r^n-1\right)}{r-1}·\frac{1}{a{r}^{n-1}}\end{array}$

Step 3: Find the value of $P^2$

$$\begin{gathered} \Rightarrow \frac{\mathrm{S}}{\mathrm{R}}=\mathrm{a}\frac{\left(1-{\mathrm{r}}^{\mathrm{n}}\right)}{1-\mathrm{r}}·\frac{\mathrm{r}-1}{{\mathrm{r}}^{\mathrm{n}}-1}\mathrm{a}·1^{\mathrm{n}-1} \\ \Rightarrow \frac{\mathrm{S}}{\mathrm{R}}=a^2·1^{(n-1)}[\mathrm{Using}(1)\&(3)] \\ \Rightarrow {\left(\frac{\mathrm{S}}{\mathrm{R}}\right)}^{\mathrm{n}}={\mathrm{a}}^{2{\mathrm{n}}_1{\mathrm{n}}^{(n-1)}} \\ \Rightarrow {\left(\frac{\mathrm{S}}{\mathrm{R}}\right)}^{\mathrm{n}}={\mathrm{p}}^2\text{[By (2)]} \end{gathered}$$

Hence, option A is correct.