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 $U_n=\sum _{n=1}^n{S}_n$, then $r{S}_n+(1-r)u_n=$

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

Answer 1

Answer: (A)

Let $r>1$. Then, $S_n=\frac{a\left(r^n-1\right)}{r-1}$
$u_n=S_1+S_2+S_3+\dots +S_n$
$=\frac{a(r-1)}{r-1}+\frac{a\left(r^2-1\right)}{r-1}+\frac{a\left(r^2-1\right)}{r-1}+\dots +\frac{a\left(r^n-1\right)}{r-1}$
$=\frac{a}{r-1}\left[\left(r+r^2+r^3+\dots +r^n\right)-n\right]$
$=\frac{a}{r-1}\left[\frac{r\left(r^n-1\right)}{r-1}-n\right]$
$\therefore r{S}_n+(1-r)u_n$
$=r\cdot \frac{a\left(r^n-1\right)}{r-1}+(1-r)\frac{\text{ar}\left(r^n-1\right)}{(1-r{)}^2}-\frac{an(1-r)}{r-1}$
$=-\frac{\text{ar}\left(r^n-1\right)}{1-r}+\frac{\text{ar}\left(r^n-1\right)}{1-r}+an=na$