A) \[\frac{n-1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}}}\]
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: A
Solution :
As given \[{{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}=.......={{a}_{n}}-{{a}_{n-1}}=d\] Where \[d\] is the common difference of the given A.P. Also\[{{a}_{n}}={{a}_{1}}+(n-1)d\]. Then by rationalising each term, \[\frac{1}{\sqrt{{{a}_{2}}}+\sqrt{{{a}_{1}}}}+\frac{1}{\sqrt{{{a}_{3}}}+\sqrt{{{a}_{2}}}}+....+\frac{1}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{n-1}}}}\] \[=\frac{\sqrt{{{a}_{2}}}-\sqrt{{{a}_{1}}}}{{{a}_{2}}-{{a}_{1}}}+\frac{\sqrt{{{a}_{3}}}-\sqrt{{{a}_{2}}}}{{{a}_{3}}-{{a}_{2}}}+.....+\frac{\sqrt{{{a}_{n}}}-\sqrt{{{a}_{n-1}}}}{{{a}_{n}}-{{a}_{n-1}}}\] \[=\frac{1}{d}\left\{ \sqrt{{{a}_{2}}}-\sqrt{{{a}_{1}}}+\sqrt{{{a}_{3}}}-\sqrt{{{a}_{2}}}+......+\sqrt{{{a}_{n}}}-\sqrt{{{a}_{n-1}}} \right\}\] \[=\frac{1}{d}\left\{ \sqrt{{{a}_{n}}}-\sqrt{{{a}_{1}}} \right\}=\frac{1}{d}\left( \frac{{{a}_{n}}-{{a}_{1}}}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}}} \right)\] \[=\frac{1}{d}\left\{ \frac{(n-1)d}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}}} \right\}=\frac{n-1}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}}}\].
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