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.
• B.
• C.
• D.

Answer 1

The correct option is D

Given that sum $S=\frac{a(r^n-1)}{r-1}=\frac{a(1-r^n)}{1-r}$ ...(i)
$P=a\left(ar\right)\left(a{r}^2\right)...\left(a{r}^{n-1}\right)=a^n{r}^{1+2+....+(n-1)}$
$=a^n{r}^{(}n-1{)}^{\frac{n}{2}}$ i.e., $P^2=a^{2n}{r}^{n(n-1)}$ ....(ii)
and $R=\frac{1}{a}+\frac{1}{ar}+\frac{1}{a{r}^2}+.....$ upto n terms
$=\frac{1}{a}(1+\frac{1}{r}+\frac{1}{r^2}+....uptonterms)$
$=\frac{\frac{1}{a}[{\left(\frac{1}{r}\right)}^n-1]}{(\frac{1}{r}-1)}(\because \frac{1}{r}>1)$ if r<1
$=\frac{(1-r^n)}{a{r}^{n-1}(1-r)}$ .....(iii)
Therefore, $\frac{S}{R}=\frac{a(1-r^n)}{1-r}\times \frac{a{r}^{n-1}(1-r)}{(1-r^n)}=a^2{r}^{n-1}$
or ${\left(\frac{S}{R}\right)}^2=(a^2{r}^{n-1}{)}^n=a^{2n}{r}^{n(n-1)}=P^2.$