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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 22

  • question_answer
    If a, b, c are in G.P. and x, y are the arithmetic means between a, b, and b, c respectively, the \[\frac{a}{x}+\frac{c}{y}\] is equal to

    A) 0                     

    B)   1

    C) 2                     

    D)   \[\frac{1}{2}\]

    Correct Answer: C

    Solution :

    Given that a, b, c are in G.P. So \[{{b}^{2}}=ac\]                                                ... (i) \[x=\frac{a+b}{2}\]                                      ? (ii) \[y=\frac{b+b}{2}\]                                      ? (iii) Now \[\frac{x}{a}+\frac{c}{y}=\frac{2a}{a+b}+\frac{2c}{b+c}=\frac{2(ab+bc+2ca)}{ab+ac+{{b}^{2}}+bc}\] \[=\,\,\frac{2(ab+bc+2ca)}{(ab+ac+ac+bc)}=2\]                        \[\left[ \because \,\,{{b}^{2}}=ac \right]\]


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