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JEE Main & Advanced JEE Main Paper (Held On 15 April 2018) Slot-I

  • question_answer
    If \[{{x}_{1}},{{x}_{2}},......,{{x}_{n}}\] and \[\frac{1}{{{h}_{1}}},\frac{1}{{{h}_{2}}},.....,\frac{1}{{{h}_{n}}}\] are two A.P.s such that \[{{x}_{3}}={{h}_{2}}=8\] and \[{{x}_{8}}={{h}_{7}}=20\], then \[{{x}_{5}}\cdot {{h}_{10}}\] equals.                                                                                                             [JEE Online 15-04-2018]

    A) 2560               

    B) 2650   

    C) 3200               

    D) 1600

    Correct Answer: A

    Solution :

    Let \[{{d}_{1}}\] be the common difference of the A.P. \[{{x}_{1}},{{x}_{2}},.....,\] then \[{{x}_{8}}-{{x}_{3}}=5{{d}_{1}}=12\Rightarrow {{d}_{1}}=\frac{12}{5}=2.4\] \[\Rightarrow {{x}_{5}}={{x}_{3}}+2{{d}_{1}}=8+2\times \frac{12}{5}=12.8\] Let \[{{d}_{2}}\] be the common difference of the other sequence then \[5{{d}_{2}}=\frac{1}{20}-\frac{1}{8}=\frac{-3}{40}\Rightarrow {{d}_{2}}=\frac{-3}{200}\] \[\Rightarrow \frac{1}{{{h}_{10}}}=\frac{1}{{{h}_{7}}}+3{{d}_{2}}=\frac{1}{200}\Rightarrow {{h}_{10}}=200\] \[\Rightarrow {{x}_{5}}\cdot {{h}_{10}}=12.8\times 200=2560\] So option A is the correct answer.


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