A) \[n-1\]
B) \[\frac{n({{a}_{1}}+{{a}_{2n}})}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{n+1}}}}\]
C) \[\frac{n-1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{n+1}}}}\]
D) None of these
Correct Answer: B
Solution :
Let\[{{a}_{1}}+{{a}_{2n}}={{a}_{2}}+{{a}_{2n-1}}=...{{a}_{n}}+{{a}_{n+1}}=k\] \[\therefore \]\[\frac{{{a}_{1}}+{{a}_{2n}}}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{2}}}}+\frac{{{a}_{2}}+{{a}_{2n-1}}}{\sqrt{{{a}_{2}}}+\sqrt{{{a}_{3}}}}+.....+\frac{{{a}_{n}}+{{a}_{n+1}}}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{n+1}}}}\] \[=k\left\{ \frac{\sqrt{{{a}_{1}}}-\sqrt{{{a}_{2}}}}{{{a}_{1}}-{{a}_{2}}}+\frac{\sqrt{{{a}_{2}}}-\sqrt{{{a}_{3}}}}{{{a}_{2}}-{{a}_{3}}}+..... \right.\] \[\left. +\frac{\sqrt{{{a}_{n}}}-\sqrt{{{a}_{n+1}}}}{{{a}_{2}}-{{a}_{n+1}}} \right\}\] \[=-\frac{k}{d}\{\sqrt{{{a}_{1}}}-\sqrt{{{a}_{2}}}+\sqrt{{{a}_{2}}}-\sqrt{{{a}_{3}}}+.....+\sqrt{{{a}_{n}}}-\sqrt{{{a}_{n+1}}}\}\]where d is common difference \[=-\frac{k}{d}\{\sqrt{{{a}_{1}}}-\sqrt{{{a}_{n+1}}}\}\] \[=\frac{k}{d}\{\sqrt{{{a}_{n+1}}}-\sqrt{{{a}_{1}}}\}\] \[=({{a}_{1}}+{{a}_{2n}}).\frac{-nd}{-d(\sqrt{{{a}_{1}}}+\sqrt{{{a}_{n+1}}})}\] \[=\frac{n({{a}_{1}}+{{a}_{2n}})}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{n+1}}}}\]
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