Question

Given a series such that the $n^{th}$ term is $[a+(n-1\left)d\right]r^{n-1}$.
If $-1<r<1$ then the sum of the infinite series is:

• A. $\frac{a}{1-r}+\frac{rd}{(1-d{)}^2}$
• B. $\frac{a}{1-r}+\frac{rd}{(1-r{)}^2}$
• C. $\frac{a}{1-d}+\frac{rd}{(1-r{)}^2}$
• D. None of these

Answer 1

Answer: (B)

$\frac{a}{1-r}+\frac{rd}{(1-r{)}^2}$

$\Rightarrow S=a+(a+d)r+(a+2d)r^2+(a+3d)r^3+.....\infty$ .....(1)

$\Rightarrow rS=ar+(a+d)r^2+(a+2d)r^3+(a+3d)r^4+.....\infty$ .....(2)

eq (1) -eq (2):

$\Rightarrow (1-r)S=a+dr+d{r}^2+d{r}^3+.....\infty$

$\Rightarrow (1-r)S=a+d.\frac{r}{1-r}$

$\Rightarrow S=\frac{a}{1-r}+\frac{rd}{(1-r{)}^2}$