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 $\operatorname{sum} \frac{1}{a_1 a_2}+\frac{1}{a_2 a_3}+\ldots . .+\frac{1}{a_n a_{n+1}}$ will approach

• A. $\frac{1}{ a _1 d }$
• B. $\frac{2}{a_1 d}$
• C. $\frac{1}{2 a _1 d }$
• 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}+\ldots . . . .+\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}+\ldots . . .+\frac{1}{a_n}-\frac{1}{a_{n+1}}\right]=\frac{1}{d}\left[\frac{1}{a_1}-\frac{1}{a_{n+1}}\right] $
$=\frac{1}{d}\left[\frac{a_{n+1}-a_1}{\left(a_1\right)\left(a_{n+1}\right)}\right]=\frac{1}{d}\left[\frac{a_1+n d-a}{\left(a_1\right)\left(a_{n+1}\right)}\right]=\frac{n}{a_1\left[a_1+n d\right]}=\frac{1}{a_1\left[\frac{a_1}{n}+d\right]}$
$\Rightarrow \text { as } n \rightarrow \infty \text { then } S=\frac{1}{a_1 d} $