Question

If $a_1,a_2,a_3,....a_n$ are in A.P. then $\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+...+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}$ =

Answer 1

$\frac{1}{\sqrt{a_1}+\sqrt{a_2}}=\frac{1}{\sqrt{a_1}+\sqrt{a_2}}\times \frac{\sqrt{a_1}-\sqrt{a_2}}{\sqrt{a_1}-\sqrt{a_2}}$
$=\frac{\sqrt{a_1}-\sqrt{a_2}}{a_1-a_2}=\frac{\sqrt{a_1}-\sqrt{a_2}}{d}$
Similarly $=\frac{1}{d}(\sqrt{a_1}-\sqrt{a_2})$
$\frac{1}{\sqrt{a_2}+\sqrt{a_3}}=\frac{\sqrt{a_2}-\sqrt{a_3}}{d}=\frac{1}{d}(\sqrt{a_2}-\sqrt{a_3})$
$\frac{1}{\sqrt{a_3}+\sqrt{a_4}}=\frac{\sqrt{a_3}-\sqrt{a_4}}{d}=\frac{1}{d}(\sqrt{a_3}-\sqrt{a_4})$
$\frac{1}{\sqrt{a_4}+\sqrt{a_5}}=\frac{\sqrt{a_4}-\sqrt{a_5}}{d}=\frac{1}{d}(\sqrt{a_4}-\sqrt{a_5})$
$\frac{1}{\sqrt{a_{n-1}}=\sqrt{a_n}}=\frac{\sqrt{a_{n-1}}-\sqrt{a_2}}{d}=\frac{1}{d}(\sqrt{a_1}-\sqrt{a_6})$
$\frac{1}{d}(\sqrt{a_1}-\sqrt{a_n})$