Question

If $S$ be the sum, $P$ the product and $R$ the sum of the reciprocals of $n$ terms of a $G.P$, prove that ${\left(S╱R\right)}^n=P^2$

Answer 1

$S=a+ar+a{r}^2+{ar}^3+..+a{r}^{n-1}$ i.e. n terms
$S=\frac{a(1-r^n)}{1-r}$
$\therefore P=Product=a.ar.a{r}^2..a{r}^{n-1}$
$=a^n{r}^{1+2+3+..n-1}=a^n{r}^{(n-1)n/2}$
$\therefore P^2=a^{2n}{r}^{n(n-1)}$ $...\left(2\right)$
$R=\frac{1}{a}+\frac{1}{ar}+\frac{1}{{ar}^2}+.....+\frac{1}{{ar}^{n-1}}$ (n terms)
$\therefore R=\frac{1}{a}.\frac{(1-\frac{1}{r^n})}{(1-\frac{1}{r})}=\frac{(r^n-1)}{r-1}.\frac{1}{{ar}^{n-1}}$
$\therefore \frac{S}{R}=a\frac{(1-r^n)}{1-r}.\frac{r-1}{r^n-1}a.r^{n-1}$
$=a^2.r^{(n-1)}$, by (1) and (3)
$\therefore {\left(\frac{S}{R}\right)}^n=a^{2n}{r}^{n(n-1)}=p^2$, by (2)
Another form:
Form (4), $\frac{S}{R}=a.{ar}^{n-1}=T_1.T_n$ tec.