Question

If S be the sum, P the product and R be the sum of the reciprocals of n terms of a GP, then $p^2$ is equal to

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

Answer 1

The correct option is D

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

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

Sum of n terms of the G.P., (S = \frac{a(r^n - 1)}{(r - 1)})

Product of n terms of the G.P.,

$P=a^{n}r\left[\frac{s(s-1)}{2}\right]$

Sum of the reciprocals of n terms of the G.P.,

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

$\therefore P^2=\left\{a^{2}r\frac{S(s-1)}{2}\right\}$

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

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

Let the first term of the G.P. be a and the common ratio be r.

Sum of n terms, $S=\frac{a(r^n-1)}{r-1}$

Product of G.P., $P=a^{n}r\frac{n(n+1)}{2}$

Sum of the reciprocals of n terms,

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

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

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