Question

Let $a_n,n\in N$ is an A.P. with common difference 'd' and all whose terms are non-zero. If $n$ approaches infinity, then the $\mathrm{sum}\frac{1}{a_1{a}_2}+\frac{1}{a_2{a}_3}+\dots ..+\frac{1}{a_n{a}_{n+1}}$ will approach

• A. $\frac{1}{a_{1}d}$
• B. $\frac{2}{a_{1}d}$
• C. $\frac{1}{2a_{1}d}$
• D. $a_{1}d$

Answer 1

Answer: (A)

$\frac{1}{d}\left[\frac{a_2-a_1}{a_1{a}_2}+\frac{a_3-a_2}{a_2{a}_3}+\dots ....+\frac{a_{n+1}-a_n}{a_n\cdot a_{n+1}}\right]$
$=\frac{1}{d}\left[\frac{1}{a_1}-\frac{1}{a_2}+\frac{1}{a_2}-\frac{1}{a_3}+\dots ...+\frac{1}{a_n}-\frac{1}{a_{n+1}}\right]=\frac{1}{d}\left[\frac{1}{a_1}-\frac{1}{a_{n+1}}\right]$
$=\frac{1}{d}\left[\frac{a_{n+1}-a_1}{\left(a_1\right)\left(a_{n+1}\right)}\right]=\frac{1}{d}\left[\frac{a_1+nd-a}{\left(a_1\right)\left(a_{n+1}\right)}\right]=\frac{n}{a_1\left[a_1+nd\right]}=\frac{1}{a_1\left[\frac{a_1}{n}+d\right]}$
$\Rightarrow \text{ as }n\to \infty \text{ then }S=\frac{1}{a_{1}d}$