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JEE Main & Advanced Mathematics Sequence & Series Question Bank Self Evaluation Test - Sequences and Series

  • question_answer
    If \[{{a}_{1}},{{a}_{2}},{{a}_{3}},.....{{a}_{n}}\] are in A.P., and\[\frac{1}{{{a}_{1}}{{a}_{n}}}+\frac{1}{{{a}_{2}}{{a}_{n-1}}}+\frac{1}{{{a}_{3}}{{a}_{n-2}}}+.....+\frac{1}{{{a}_{n}}{{a}_{1}}}\]\[=K\left( \frac{1}{{{a}_{1}}}+\frac{1}{{{a}_{2}}}+\frac{1}{{{a}_{3}}}+....+\frac{1}{{{a}_{n}}} \right)\]. Then K is

    A) \[\frac{2}{{{a}_{1}}+{{a}_{n}}}\]

    B) \[\frac{n}{{{a}_{1}}+{{a}_{n}}}\]

    C) \[\frac{1}{{{a}_{1}}+{{a}_{n}}}\]

    D) \[\frac{n-1}{{{a}_{1}}+{{a}_{n}}}\]

    Correct Answer: A

    Solution :

    [a] We know that in an A.P. \[{{a}_{1}}+{{a}_{n}}={{a}_{2}}+{{a}_{n-1}}={{a}_{3}}+{{a}_{n-2}}=\]    ???.. (i) [see the properties of A.P.] \[\therefore \,\,\frac{1}{{{a}_{1}}{{a}_{n}}}+\frac{1}{{{a}_{2}}{{a}_{n-1}}}+\frac{1}{{{a}_{3}}{{a}_{n-2}}}+....+\frac{1}{{{a}_{n}}{{a}_{1}}}\] \[=\frac{1}{{{a}_{1}}+{{a}_{n}}}\left[ \frac{{{a}_{1}}+{{a}_{n}}}{{{a}_{1}}{{a}_{n}}}+\frac{{{a}_{1}}+{{a}_{n}}}{{{a}_{2}}{{a}_{n-2}}}+\frac{{{a}_{1}}+{{a}_{n}}}{{{a}_{3}}{{a}_{n-2}}}+....+\frac{{{a}_{1}}+{{a}_{n}}}{{{a}_{n}}{{a}_{1}}} \right]\]\[=\frac{2}{{{a}_{1}}+{{a}_{n}}}\left[ \frac{1}{{{a}_{1}}}+\frac{1}{{{a}_{2}}}+\frac{1}{{{a}_{3}}}+......+\frac{1}{{{a}_{n}}} \right]\]


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