Question

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

• A. na
• B. n
• C. $n^{2}a$
• D. None of these

Answer 1

The correct option is A na

Let $r<1,$ then $S_n=\frac{a(l-r^n)}{1-r}Now,U_n=S_1+S_2+S_3+\cdots +S_n=\frac{a(1-r)}{1-r}+\frac{a(1-r^2)}{1-r}+\frac{a(1-r^3)}{1-r}+\cdots +\frac{a(1-r^n)}{1-r}=\frac{a}{1-r}\left[\right(1+1+1+\cdots ntimes)-(r+r^2+r^3+\cdots +r^n\left)\right]=\frac{a}{1-r}[n-\frac{r(1-r^n)}{1-r}]\therefore (1-r)U_n=a[n-\frac{r(1-r^n)}{1-r}]=an-\frac{ar(1-r^n)}{1-r}=an-r{S}_n$
$\therefore r{S}_n+(1-r)U_n=na$