Question

In a geometric series, the first term is $a$ and common ratio is $r .$ If $S _{n}$ denotes the sum of $n$ terms and $U_{n}=\displaystyle\sum_{n=1}^{n} S _{n}$, then $r S_{n}+(1-r) u_{n}=$

• A. $n a$
• B. $(n-1) a$
• C. $(n+1) a$
• D. None of these

Answer 1

Answer: (A)

Let $r>1$. Then, $S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}$
$u_{n}=S_{1}+S_{2}+S_{3}+\ldots+S_{n}$
$=\frac{a(r-1)}{r-1}+\frac{a\left(r^{2}-1\right)}{r-1}+\frac{a\left(r^{2}-1\right)}{r-1}+\ldots+\frac{a\left(r^{n}-1\right)}{r-1}$
$=\frac{a}{r-1}\left[\left(r+r^{2}+r^{3}+\ldots+r^{n}\right)-n\right]$
$=\frac{a}{r-1}\left[\frac{r\left(r^{n}-1\right)}{r-1}-n\right]$
$\therefore r S_{n}+(1-r) u_{n}$
$=r \cdot \frac{a\left(r^{n}-1\right)}{r-1}+(1-r) \frac{\text{ar}\left(r^{n}-1\right)}{(1-r)^{2}}-\frac{a n(1-r)}{r-1}$
$=-\frac{\text{ar}\left(r^{n}-1\right)}{1-r}+\frac{\text{ar}\left(r^{n}-1\right)}{1-r}+a n=n a$