Question

If, in a $G.P.$ of $3n$ terms, $S_1$ denotes the sum of the first $n$ terms, $S_2$ the sum of the second block of $n$ terms and $S_3$ the sum of the last $n$ terms, then $S_1,S_2,S_3$ are in

• A. $A.P.$
• B. $G.P.$
• C. $H.P.$
• D. $A.G.P.$

Answer 1

The correct option is B $G.P.$

Let the $3n$ terms of $G.P.$ be
$a,ar,a{r}^2,\dots ,a{r}^{n-1},a{r}^n,a{r}^{n+1},a{r}^{n+2},\dots ,a{r}^{2n-1},a{r}^{2n},a{r}^{2n+1},a{r}^{2n+2},\dots ,a{r}^{3n-1}$

Then,
$S_1=a+ar+a{r}^2+\dots +a{r}^{n-1}=\frac{a(1-r^n)}{1-r}$
$S_2=a{r}^n+a{r}^{n+1}+a{r}^{n+2}+\dots +a{r}^{2n-1}=\frac{a{r}^n(1-r^n)}{1-r}$
$S_3=a{r}^{2n}+a{r}^{2n+1}+a{r}^{2n+2}+\dots +a{r}^{3n-1}=\frac{a{r}^{2n}(1-r^n)}{1-r}$

Now, $S_2^2=\frac{a^2{r}^{2n}(1-r^n{)}^2}{(1-r{)}^2}$
$\Rightarrow S_2^2=\frac{a(1-r^n)}{(1-r)}\times \frac{a{r}^{2n}(1-r^n)}{(1-r)}\Rightarrow S_2^2=S_1\times S_3$
Hence, $S_1,S_2,S_3$ are in $G.P.$