Question

If $ {{a}_{1}},{{a}_{2}},{{a}_{3}},...,{{a}_{n}} $ are in AP, where $ {{a}_{i}}>0 $ for all $ i $ , then value of the expression $ \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. $ \frac{n-1}{\sqrt{{{a}_{n}}}-\sqrt{{{a}_{1}}}} $
• B. $ \frac{n-1}{\sqrt{{{a}_{1}}}-\sqrt{{{a}_{n}}}} $
• C. $ \frac{n-1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}}} $
• D. None of these

Answer 1

Answer: (C)

Let d be the common difference of the given AP then, $ {{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}={{a}_{4}}-{{a}_{3}} $ $ =...={{a}_{n}}-{{a}_{n-1}}=d $ ?(i) Now, $ \frac{1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{2}}}}+\frac{1}{\sqrt{{{a}_{2}}}+\sqrt{{{a}_{3}}}}+\frac{1}{\sqrt{{{a}_{3}}}+\sqrt{{{a}_{4}}}} $ $ +....+\frac{1}{\sqrt{{{a}_{n-1}}}+\sqrt{{{a}_{n}}}} $ ?(ii) On multiplying by the rationalisation factor of every term in numerator and denominator, we get $ \frac{\sqrt{{{a}_{2}}}-\sqrt{{{a}_{1}}}}{{{a}_{2}}-{{a}_{1}}}+\frac{\sqrt{{{a}_{3}}}-\sqrt{{{a}_{2}}}}{{{a}_{3}}-{{a}_{2}}}+\frac{\sqrt{{{a}_{4}}}-\sqrt{{{a}_{3}}}}{{{a}_{4}}-{{a}_{3}}} $ $ +....+\frac{\sqrt{{{a}_{n}}}-\sqrt{{{a}_{n-1}}}}{{{a}_{n}}-{{a}_{n-1}}} $ [using Eq.(i)] $ =\frac{1}{d}[\sqrt{{{a}_{2}}}-\sqrt{{{a}_{1}}}+\sqrt{{{a}_{3}}}-\sqrt{{{a}_{2}}}+\sqrt{{{a}_{4}}}-\sqrt{{{a}_{3}}} $ $ +...+\sqrt{{{a}_{n}}}-\sqrt{{{a}_{n-1}}}] $ $ =\frac{1}{d}(\sqrt{{{a}_{n}}}-\sqrt{{{a}_{1}}}) $ $ =\frac{1}{d}(\sqrt{{{a}_{n}}}-\sqrt{{{a}_{1}}}).\frac{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}}}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}}} $ $ =\frac{{{a}_{n}}-{{a}_{1}}}{d(\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}})} $ $ =\frac{n-1}{d(\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}})} $ $ [\because {{a}_{n}}={{a}_{1}}+(n-1)d] $ $ =\frac{n-1}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}}} $ $ =\frac{n-1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}}} $