Question

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

• 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. $n+1$

Answer 1

Answer: (B)

$\frac{n(a_1+a_{2n})}{\sqrt{a_1}+\sqrt{a_{n+1}}}$

$a_1,a_2,a_3,.....a_{2n}$ are in A.P.
Let $d$ be the common difference, then
$S=\frac{2n}{2}(a_1+a_{2n})$
$\Rightarrow S=n(a_1+a_{2n})$
Consider ,$\frac{a_1+a_{2n}}{\sqrt{a_1}+\sqrt{a_2}}+\frac{a_2+a_{2n-1}}{\sqrt{a_2}+\sqrt{a_3}}+\frac{a_3+a_{2n-2}}{\sqrt{a_3}+\sqrt{a_4}}+...+\frac{a_n+a_{n+1}}{\sqrt{a_n}+\sqrt{a_{n+1}}}$
$=\frac{a_1+a_{2n}}{\sqrt{a_1}+\sqrt{a_2}}\frac{\sqrt{a_1}-\sqrt{a_2}}{\sqrt{a_1}-\sqrt{a_2}}+\frac{a_2+a_{2n-1}}{\sqrt{a_2}+\sqrt{a_3}}\frac{\sqrt{a_2}-\sqrt{a_3}}{\sqrt{a_2}-\sqrt{a_3}}+...+\frac{a_n+a_{n+1}}{\sqrt{a_n}+\sqrt{a_{n+1}}}\frac{\sqrt{a_n}-\sqrt{a_{n+1}}}{\sqrt{a_n}-\sqrt{a_{n+1}}}$
$=\frac{S(\sqrt{a_1}-\sqrt{a_2})}{-nd}+\frac{S(\sqrt{a_2}-\sqrt{a_3})}{-nd}+...+\frac{S(\sqrt{a_n}-\sqrt{a_{n+1}})}{-nd}$
$=\frac{S}{-nd}(\sqrt{a_1}-\sqrt{a_2}+\sqrt{a_2}-\sqrt{a_3}+.......-\sqrt{a_n}+\sqrt{a_n}-\sqrt{a_{n+1}})$
$=\frac{S}{-nd}(\sqrt{a_1}-\sqrt{a_{n+1}})$
$=\frac{(a_1+a_{2n})}{d}(\sqrt{a_{n+1}}-\sqrt{a_1})$
$=\frac{(a_1+a_{2n})}{d(\sqrt{a_{n+1}}-\sqrt{a_1})}(a_{n+1}-a_1)$
$=\frac{(a_1+a_{2n})}{d(\sqrt{a_{n+1}}+\sqrt{a_1})}nd$
$=\frac{n(a_1+a_{2n})}{(\sqrt{a_{n+1}}+\sqrt{a_1})}$