Question

If the sum of the $n$ terms of the G.P is $S$ product is $P$ and sum of their inverse is $R$, then $P^2$ is equal to

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

Answer 1

The correct option is D ${\left(\frac{S}{R}\right)}^n$

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

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

$P=a.ar.a{r}^{2}a{r}^3.....a{r}^{n-1}$

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

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

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

$R=\frac{1}{a}+\frac{1}{ar}+\frac{1}{a{r}^2}+\frac{1}{a{r}^3}......+\frac{1}{a{r}^{n-1}}$

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

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

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

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

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