Question

If $a_r>0,r\in N$ and $a_1,a_2,a_3,...,a_{2n}$ are in A.P. then $\frac{a_1+a_{2n}}{\sqrt{a_1}+\sqrt{a_2}}+\frac{a_2+a_{2n-1}}{\sqrt{a_2}+\sqrt{a_3}}+\frac{a_3+a_{2n-2}}{\sqrt{a_3}+\sqrt{a_4}}+...+\frac{a_n+a_{n+1}}{\sqrt{a_n}+\sqrt{a_{n+1}}}$ is equal to

• A. $n-1$
• B. $\frac{n(a_1+a_{2n})}{\sqrt{a_1}+\sqrt{a_{n+1}}}$
• C. $\frac{n-1}{\sqrt{a_1}+\sqrt{a_{n+1}}}$
• D. None of these

Answer 1

The correct option is B $\frac{n(a_1+a_{2n})}{\sqrt{a_1}+\sqrt{a_{n+1}}}$

$a_1+a_{2n}=a_2+a_{2n-1}=...=a_n+a_{n+1}=k($say$)$
Expression $=k(\frac{\sqrt{a_1}-\sqrt{a_2}}{a_1-a_2}+...+\frac{\sqrt{a_n}-\sqrt{a_{n+1}}}{a_n-a_{n+1}})$
$=\frac{k}{-d}(\sqrt{a_1}-\sqrt{a_{n+1}})$, where $d=$ common difference
$=\frac{k}{-d}\frac{a_1-a_{n+1}}{\sqrt{a_1}+\sqrt{a_{n+1}}}=(a_1+a_{2n}).\frac{-nd}{-d(\sqrt{a_1}+\sqrt{a_{n+1}})}$