Question

If $a_1,a_2,.....,a_n$ On are in arithmetic progression, where $a_i>0$ for all $i$ . Then $\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+...+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}$ is equal to

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

Answer 1

Answer: (B)

Since, $a_1,a_2,...,a_n$ are in AP. Then, $a_2-a_1=a_3-a_2=.....=a_n-a_{n-1}=d$ where d is common difference Now, $\frac{1}{\sqrt{a_2}+\sqrt{a_1}}+\frac{1}{\sqrt{a_3}+\sqrt{a_2}}$ $+......+\frac{1}{\sqrt{a_n}+\sqrt{a_{n-1}}}$ $=\frac{\sqrt{a_2}-\sqrt{a_1}}{d}+\frac{\sqrt{a_3}-\sqrt{a_2}}{d}$ $+...+\frac{\sqrt{a_n}-\sqrt{a_{n-1}}}{d}$ $=\frac{1}{d}(\sqrt{a_n}-\sqrt{a_1})\times \frac{\sqrt{a_n}+\sqrt{a_1}}{\sqrt{a_n}+\sqrt{a_1}}$ $=\frac{1}{d}\left(\frac{a_n-a_1}{\sqrt{a_n}+\sqrt{a_1}}\right)=\frac{n-1}{\sqrt{a_n}+\sqrt{a_1}}$