Question

If S be the sum, P the product and R the sum of the reciprocals of n terms of a G.P., then ${\left(\frac{S}{R}\right)}^n$ =

• A. P
• B. P²
• C. P³
• D. √P

Answer 1

The correct option is B

P²

S = a + ar + a $r^2$ + a $r^2$ + ......+ a $r^{n-1}$

i.e. n terms

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

∴ P = product = a.ar.a $r^2$. .........a $r^{n-1}$ = $a^n$ $r^{1+2+3+4+5+.....+n-1}$ = $a^n$ . $r^{\frac{n(n-1)}{2}}$

∴ $p^2$ = $a^{2n}$ $r^{n(n-1)}$ ......(ii)

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

∴ R = $\frac{1}{a}$ $\frac{(1-\frac{1}{r^n})}{1-\frac{1}{r}}$ = $\frac{(r^n-1)}{r-1}$. $\frac{1}{a{r}^{n-1}}$ .........(iii)

∴ $\frac{S}{R}$ = a. $\frac{1-r^n}{1-r}$. $\frac{(r-1)}{(r^n-1)}$.a $r^{n-1}$ = $a^2$ $r^{(n-1)}$ by (i) and (ii)

∴ ${\left[\frac{S}{R}\right]}^n$ = $a^{2n}$ $r^{n(n-1)}$ = $p^2$ by (ii)