Question

If $a_1,a_2,.....,a_n$are in AP, then $\frac{1}{a_1{a}_2}+\frac{1}{a_2{a}_3}+...\dots .+\frac{1}{a_{(n-1)}{a}_n}$ is

• A. $\frac{(n-1)}{a_1{a}_n}$
• B. $\frac{1}{a_1{a}_n}$
• C. $\frac{(n+1)}{a_1{a}_n}$
• D. $\frac{n}{a_1{a}_n}$

Answer 1

The correct option is A

$\frac{(n-1)}{a_1{a}_n}$

Find the value of $\frac{1}{a_1{a}_2}+\frac{1}{a_2{a}_3}+...\dots .+\frac{1}{a_{(n-1)}{a}_n}$:

Given $a_1,a_2,.....,a_n$are in AP.

Let $d$ be the common difference of AP.

$\begin{array}{rcl}d& =& a_2-a_1=a_3-a_2=a_4-a_3=...\dots .=a_n-a_{(n-1)}\\ \frac{1}{a_1{a}_2}+\frac{1}{a_2{a}_3}+....................+\frac{1}{a_{(n-1)}{a}_n}& =& \frac{1}{d}\left[\frac{d}{a_1{a}_2}+\frac{d}{a_2{a}_3}+....................+\frac{d}{a_{(n-1)}{a}_n}\right]\\ & =& \frac{1}{d}\left[\frac{(a_2-a_1)}{a_1{a}_2}+\frac{(a_3-a_2)}{a_2{a}_3}+....................+\frac{(a_n-a_{(n-1)})}{a_{(n-1)}{a}_n}\right]\\ & =& \frac{1}{d}\left[\frac{a_2}{a_1{a}_2}-\frac{a_1}{a_1{a}_2}+\frac{a_3}{a_2{a}_3}-\frac{a_2}{a_2{a}_3}+....................+\frac{a_n}{a_{(n-1)}{a}_n}-\frac{a_{(n-1)}}{a_{(n-1)}{a}_n}\right]\\ & =& \frac{1}{d}\left[\frac{1}{a_1}-\frac{1}{a_2}+\frac{1}{a_2}-\frac{1}{a_3}+....................+\frac{1}{a_{(n-1)}}-\frac{1}{a_n}\right]\\ & =& \frac{1}{d}\left[\frac{1}{a_1}-\frac{1}{a_n}\right]\\ & =& \frac{1}{d}\left[\frac{(a_n-a_1)}{a_1{a}_n}\right]\\ & =& \frac{1}{d}\left[\frac{a_1+(n-1)d-a_1}{a_1{a}_n}\right]\mathbf{\left[}\mathbf{\because }{\mathbf{a}}_{\mathbf{n}}\mathbf{=}{\mathbf{a}}_{\mathbf{1}}\mathbf{+}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{d}\mathbf{}\mathbf{\right]}\\ & =& \frac{1}{d}\left[\frac{(n-1)d}{a_1{a}_n}\right]\\ & =& \frac{(n-1)}{a_1{a}_n}\end{array}$

Hence, the correct option is (A).