Question

In a G.P., the first term and common ratio is $a$ and $r$ respectively. If $S_n$ denotes the sum of $n$ terms and $U_n=\sum _{n=1}^n{S}_n$, then $r{S}_n+(1-r)U_n$ is equal to

• A. $an$
• B. $(n-1)a$
• C. $(n+1)a$
• D. $a$

Answer 1

The correct option is A $an$

We know that,
$S_n=\frac{a(r^n-1)}{r-1}$
Now,
$U_n=\sum _{n=1}^n{S}_n\Rightarrow U_n=\sum _{n=1}^n\frac{a(r^n-1)}{r-1}\Rightarrow U_n=\frac{a}{r-1}\sum _{n=1}^n(r^n-1)\Rightarrow U_n=\frac{a}{r-1}\left[\sum _{n=1}^n{r}^n-\sum _{n=1}^{n}1\right]\Rightarrow U_n=\frac{a}{r-1}[\frac{r(r^n-1)}{r-1}-n]$

Now,
$r{S}_n+(1-r)U_n=r\cdot \frac{a(r^n-1)}{r-1}+(1-r)\cdot [\frac{ar(r^n-1)}{(r-1{)}^2}-\frac{an}{r-1}]=\frac{ar(r^n-1)}{r-1}-\frac{ar(r^n-1)}{r-1}+an=an$