Question

For first $n$ natural numbers we have the following results with usual notations $\sum _{r=1}^{n}r=\frac{n(n+1)}{2},\sum _{r=1}^n{r}^2=\frac{n(n+1)(2n+1)}{6},\sum _{r=1}^n{r}^3={\left(\sum _{r=1}^{n}r\right)}^2$ If $a_1{a}_2....a_n\in A.P$ then sum to $n$ terms of the sequence $\frac{1}{a_1{a}_2},\frac{1}{a_2{a}_3},...\frac{1}{a_{n-1}{a}_n}$ is equal to $\frac{n-1}{a_1{a}_n}$
and the sum to $n$ terms of a $G.P$ with first term '$a$' & common ratio '$r$' is given by $S_n=\frac{lr-a}{r-1}$ for $r\ne 1$ for $r=1$ sum to $n$ terms of same $G.P.$ is $n$ $a$, where the sum to infinite terms of$G.P.$ is the limiting value of
$\frac{lr-a}{r-1}$ when $n\to \infty ,\left|r\right|<l$ where $l$ is the last term of $G.P.$ On the basis of above data answer the following questionsThe sum to infinite terms of the series $\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+..$ is equal to ?

• A. $\frac{4}{3}$
• B. $\frac{3}{4}$
• C. $\frac{8}{3}$
• D. Does not exit

Answer 1

The correct option is B $\frac{3}{4}$

Let, $S_{\infty }=\frac{1}{2}+\frac{1}{6}+d\frac{1}{18}+..\infty$
$\Rightarrow S_{\infty }=\frac{1}{2}(1+\frac{1}{3}+\frac{1}{3^2}+....\infty )$
As we know that, sum of infinite G.P series $=\frac{a}{1-r}$
Therefore, $S_{\infty }=\frac{1}{2}\left(\frac{1}{1-\left(1/3\right)}\right)=\frac{3}{4}$
Ans: B