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
Correct Answer: B
Solution :
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, \[\frac{1}{\sqrt{{{a}_{2}}}+\sqrt{{{a}_{1}}}}+\frac{1}{\sqrt{{{a}_{3}}}+\sqrt{{{a}_{2}}}}\] \[+......+\frac{1}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{n-1}}}}\] \[=\frac{\sqrt{{{a}_{2}}}-\sqrt{{{a}_{1}}}}{d}+\frac{\sqrt{{{a}_{3}}}-\sqrt{{{a}_{2}}}}{d}\] \[+...+\frac{\sqrt{{{a}_{n}}}-\sqrt{{{a}_{n-1}}}}{d}\] \[=\frac{1}{d}(\sqrt{{{a}_{n}}}-\sqrt{{{a}_{1}}})\times \frac{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}}}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}}}\] \[=\frac{1}{d}\left( \frac{{{a}_{n}}-{{a}_{1}}}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}}} \right)=\frac{n-1}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}}}\]You need to login to perform this action.
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