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 = $\frac{a(r^{\pi }-1)}{r-1}$ = $\frac{a(1-r^{\pi })}{(1-r)}$ ..........(i)

P = a(ar)(a $r^2$) ..........(a $r^{\pi -1}$) = $a^{\pi }$ $r^{1+2+--------+(\pi -1)}$

= $a^{\pi }$ $r^{\frac{(\pi -1)\pi }{2}}$ i.e., $P^2$ = $a^{2\pi }$ $r^{\pi (\pi -1)}$ ...........(ii)

and R = $\frac{1}{a}$ + $\frac{1}{ar}$ + $\frac{1}{a{r}^2}$ + .........upto n terms

= $\frac{1}{a}$ ( 1 + $\frac{1}{r}$ + $\frac{1}{r^2}$ + ...........upto n terms)

= $\frac{\frac{1}{a}[\frac{\frac{1}{r}}{n}-1]}{(\frac{1}{r}-1)}$ ( ∵ $\frac{1}{r}$ > 1) if r < 1

= $\frac{(1-r^{\pi })}{a{r}^{\pi -1}(1-r)}$ --------(iii)

Therefore, $\frac{S}{R}$ = $\frac{a(1-r^{\pi })}{1-r}$ × $\frac{a{r}^{\pi -1}(1-r)}{(1-r^{\pi })}$ = $a^2$ ${a^2}^{\pi -1}$

or ${\left(\frac{S}{R}\right)}^{\pi }$ = ${a^2{r}^{\pi -1}}^{\pi }$ = $a^{2\pi }$ $r^{\pi (\pi -1)}$ = $P^2$.