Question

Let $ a_1, a_2 ,a_3 ....,a_n $ be in AP. If $ \frac{1}{a_1a_n}+\frac{1}{a_2a_{n-1}}+....+\frac{1}{a_na_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
$a_{1}+a_{n}=a_{2}+a_{n-1}$
$=a_{3}+a_{n-2}=\dots =a_{r}+a_{n-(r-1)}$
Now, $\frac{1}{a_{1}a_{n}}+\frac{1}{a_{2} a_{n-1}}+\frac{1}{a_{3} a_{n-3}}+\ldots+\frac{1}{a_{n} a_{1}} $
$=\frac{1}{a_{n}+a_{1}}\left[\frac{a_{n}+a_{1}}{a_{1} a_{n}}+\frac{a_{n} + a_{1}}{a_{2} a_{n-1}}+\ldots+\frac{a_{n}+a_{1}}{a_{n}a_{1}}\right]$
$=\frac{1}{a_{n}+a_{1}}\left[\frac{a_{n}+a_{1}}{a_{1}a_{n}}+\frac{a_{n-1}+a_{2}}{a_{2}a_{n-1}}+\ldots+\frac{a_{n}+a_{1}}{a_{n}a_{1}}\right]$
$=\frac{1}{a_{n} +a_{1}}\left[\frac{1}{a_{n}}+\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{n-1}}+\ldots+\frac{1}{a_{n}}+\frac{1}{a_{1}}\right]$
$=\frac{1}{a_{n}+a_{1}}\left[\frac{2}{a_{1}}+\frac{2}{a_{2}}+\frac{2}{a_{3}}+\ldots+\frac{2}{a_{n}}\right]$
$=\frac{2}{a_{n}+a_{1}}\left[\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\ldots+\frac{1}{a_{n}}\right]$
$\therefore k=2$