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}}}} $