Question

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

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

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