Question

Let there be $n$ numbers in G.P. whose common ratio is $r$ and $S_m$ denotes the sum of their first $m$ terms. The sum of their products taken two at a time is $k{S}_n{S}_{n-1}$ where $k=$

• A. $\frac{r-1}{r}$
• B. $\frac{r-1}{r+1}$
• C. $\frac{r}{r+1}$
• D. None of these

Answer 1

Answer: (C)

Let the $n$ numbers in G.P. be $a,ar,a{r}^2,\dots ,a{r}^{n-1}$
Thus, we have,
$S_n=a\left(\frac{1-r^n}{1-r}\right)$
Also,
$S_n^2={\left[a+ar+a{r}^2+\dots +a{r}^{n-1}\right]}^2$
$\Rightarrow a^2\left(\frac{1-r^n}{1-r}\right)=a^2+(ar{)}^2+{\left(a{r}^2\right)}^2+\dots ..+{\left(a{r}^{n-1}\right)}^2+2S$
where $S$ denotes the sum of the product of the terms of the G.P. taken two at a time.
$\Rightarrow a^2{\left(\frac{1-r^n}{1-r}\right)}^2=a^2\left(\frac{1-r^{2n}}{1-r^2}\right)+2S$
$\Rightarrow S=\frac{a^2}{2}\left[\frac{{\left(1-r^n\right)}^2}{(1-r{)}^2}-\frac{1-r^{2n}}{1-r^2}\right]$
$=\frac{a^2}{2}\left(\frac{1-r^n}{1-r}\right)\left[\frac{1-r^n}{1-r}-\frac{1+r^n}{1+r}\right]$
$=S_n\times \frac{a}{2}\times \frac{\left(1-r^n\right)(1+r)-\left(1+r^n\right)(1-r)}{(1+r)(1-r)}$
$=S_n\times \frac{a}{2}\times \frac{2\left(r-r^n\right)}{(1+r)(1-r)}$
$=S_n\times \frac{r}{1+r}\times \frac{a\left(1-r^{n-1}\right)}{1-r}$
$=\left(\frac{r}{1+r}\right)S_n{S}_{n-1}$
$\therefore k=\frac{r}{r+1}$