Question

Let $a_1,a_2,a_3....,a_n$ be in AP. If $\frac{1}{a_1{a}_n}+\frac{1}{a_2{a}_{n-1}}+....+\frac{1}{a_n{a}_1}$ = $\frac{K}{a_1+a_n}[\frac{1}{a_1}+\frac{1}{a_2}+....+\frac{1}{a_n}]$ then $K$ is equal to

• A. 1
• B. 2
• C. 3
• D. 5

Answer 1

Answer: (B)

We have,
$a_1,a_2,a_3,a_4\dots ,a_n$ are in AP.
We know that, if $a_1,a_2,a_3\dots ,a_n$ in $AP$, then
$a_1+a_n=a_2+a_{n-1}$
$=a_3+a_{n-2}=\cdots =a_r+a_{n-(r-1)}$
Now, $\frac{1}{a_1{a}_n}+\frac{1}{a_2{a}_{n-1}}+\frac{1}{a_3{a}_{n-3}}+\dots +\frac{1}{a_n{a}_1}$
$=\frac{1}{a_n+a_1}\left[\frac{a_n+a_1}{a_1{a}_n}+\frac{a_n+a_1}{a_2{a}_{n-1}}+\dots +\frac{a_n+a_1}{a_n{a}_1}\right]$
$=\frac{1}{a_n+a_1}\left[\frac{a_n+a_1}{a_1{a}_n}+\frac{a_{n-1}+a_2}{a_2{a}_{n-1}}+\dots +\frac{a_n+a_1}{a_n{a}_1}\right]$
$=\frac{1}{a_n+a_1}\left[\frac{1}{a_n}+\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_{n-1}}+\dots +\frac{1}{a_n}+\frac{1}{a_1}\right]$
$=\frac{1}{a_n+a_1}\left[\frac{2}{a_1}+\frac{2}{a_2}+\frac{2}{a_3}+\dots +\frac{2}{a_n}\right]$
$=\frac{2}{a_n+a_1}\left[\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\dots +\frac{1}{a_n}\right]$
$\therefore k=2$