Question

$a_1,a_2,\dots \dots .a_n$ are A.P. Where $a_1$ > 0 for all i, then $\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+....+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Let d be the common difference

$\Rightarrow \frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+....+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}$

$\Rightarrow \frac{\sqrt{a_2}-\sqrt{a_1}}{d}+\frac{\sqrt{a_3}-\sqrt{a_2}}{d}+...+\frac{\sqrt{a_n}-\sqrt{a_{n-1}}}{d}=\frac{\sqrt{a_n}-\sqrt{a_1}}{d}$ , cancelling the terms

$\frac{\sqrt{a_2}-\sqrt{a_1}}{d}+\frac{\sqrt{a_3}-\sqrt{a_2}}{d}+...+\frac{\sqrt{a_n}-\sqrt{a_{n-1}}}{d}=\frac{\sqrt{a_n}-\sqrt{a_1}}{d}=\frac{a_n-a_1}{(\sqrt{a_1}+\sqrt{a_2})d}=\frac{(n-1)}{\sqrt{a_n}+\sqrt{a_1}}$