Question

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

• A. $P$
• B. $P^2$
• C. $P^{2n}$
• D. $P^n$

Answer 1

The correct option is B $P^2$

Let $a$ be the first term and $r$ the common ratio of $G.P.$
$S=a+ar+a{r}^2+...\text{to}n\text{terms}$
$\therefore S=\frac{a(1-r^n)}{1-r}$

Now,
$P=a\cdot ar\cdot a{r}^2\cdot a{r}^3...\text{to}n\text{terms}$
$=a^n\cdot r^{1+2+3+...+(n-1)}=a^n\cdot r^{\frac{(n-1)n}{2}}$

Again,
$R=\frac{1}{a}+\frac{1}{ar}+\frac{1}{a{r}^2}+\frac{1}{a{r}^3}+...\text{to}n\text{terms}$
$=\frac{\frac{1}{a}[1-{\left(\frac{1}{r}\right)}^n]}{1-\frac{1}{r}}=\frac{1}{a}\cdot \frac{(r^n-1)}{r^n}\cdot \frac{r}{(r-1)}$
$\therefore R=\frac{r^n-1}{r-1}\cdot \frac{1}{a{r}^{n-1}}=\frac{1-r^n}{1-r}\cdot \frac{1}{a{r}^{n-1}}$

Now,
$\frac{S}{R}=\frac{a(1-r^n)}{1-r}\cdot \frac{(1-r)a{r}^{n-1}}{1-r^n}=a^2{r}^{n-1}$

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