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