A) \[\frac{n-1}{{{a}_{1}}{{a}_{n+1}}}\]
B) \[\frac{1}{{{a}_{1}}{{a}_{n+1}}}\]
C) \[\frac{n+1}{{{a}_{1}}{{a}_{n+1}}}\]
D) \[\frac{n}{{{a}_{1}}{{a}_{n+1}}}\]
Correct Answer: D
Solution :
\[{{a}_{1}},{{a}_{2}},{{a}_{3}},.......,{{a}_{n+1}}\] are in A.P. and common difference = d Let \[S=\frac{1}{{{a}_{1}}{{a}_{2}}}+\frac{1}{{{a}_{2}}{{a}_{3}}}+..........+\frac{1}{{{a}_{n}}{{a}_{n+1}}}\] Þ \[S=\frac{1}{d}\,\left\{ \frac{d}{{{a}_{1}}{{a}_{2}}}+\frac{d}{{{a}_{2}}{{a}_{3}}}+......+\frac{d}{{{a}_{n}}\,\,{{a}_{n+1}}} \right\}\] Þ \[S=\frac{1}{d}\,\left\{ \frac{{{a}_{2}}-{{a}_{1}}}{{{a}_{1}}{{a}_{2}}}+\frac{{{a}_{3}}-{{a}_{2}}}{{{a}_{2}}{{a}_{3}}}+......+\frac{{{a}_{n+1}}-{{a}_{n}}}{{{a}_{n}}\,\,\,{{a}_{n+1}}} \right\}\] Þ \[S=\frac{1}{d}\left\{ \frac{1}{{{a}_{1}}}-\frac{1}{{{a}_{2}}}+\frac{1}{{{a}_{2}}}-\frac{1}{{{a}_{3}}}+.......+\frac{1}{{{a}_{n}}}-\frac{1}{{{a}_{n+1}}} \right\}\] Þ \[S=\frac{1}{d}\left\{ \frac{1}{{{a}_{n}}}-\frac{1}{{{a}_{n+1}}} \right\}=\frac{1}{d}\left\{ \frac{{{a}_{n+1}}-{{a}_{1}}}{{{a}_{1}}{{a}_{n+1}}} \right\}\] Þ \[S=\frac{1}{d}\left( \frac{nd}{{{a}_{1}}{{a}_{n+1}}} \right)=\frac{n}{{{a}_{1}}{{a}_{n+1}}}\]. Trick: Check for \[n=2\].
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