A) \[\frac{n+1}{\sqrt{{{a}_{1}}+{{a}_{n}}}}\]
B) \[\frac{n-1}{\sqrt{{{a}_{1}}}-\sqrt{{{a}_{n}}}}\]
C) \[\frac{n+1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}}}\]
D) \[\frac{n-1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}}}\]
Correct Answer: D
Solution :
Since,\[{{a}_{1}},\,\,{{a}_{2}},...,{{a}_{n}}\]are in\[AP\]. \[\therefore \]\[{{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}=...={{a}_{n}}-{{a}_{n-1}}=d\] (say) where, \[d\] is a common difference of an\[AP\]. Now, \[\frac{1}{\sqrt{{{a}_{2}}}+\sqrt{{{a}_{1}}}}+\frac{1}{\sqrt{{{a}_{3}}}+\sqrt{{{a}_{2}}}}+...+\frac{1}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{n-1}}}}\] \[=\frac{\sqrt{{{a}_{2}}}-\sqrt{{{a}_{1}}}}{{{a}_{2}}-{{a}_{1}}}+\frac{\sqrt{{{a}_{3}}}-\sqrt{{{a}_{2}}}}{{{a}_{3}}-{{a}_{2}}}+...+\frac{\sqrt{{{a}_{n}}}-\sqrt{{{a}_{n-1}}}}{{{a}_{n}}-{{a}_{n-1}}}\] \[=\frac{1}{d}(\sqrt{{{a}_{n}}}-\sqrt{{{a}_{2}}})=\frac{{{a}_{n}}-{{a}_{1}}}{d(\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}})}\] \[=\frac{1}{d}\left( \frac{(n-1)d}{\sqrt{{{a}_{n}}}-\sqrt{{{a}_{1}}}} \right)=\frac{n-1}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}}}\]
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