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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    If a, b, c are in GP and\[x,y\]are arithmetic mean of a, b and b, c respectively, then \[\frac{1}{x}+\frac{1}{y}\] is equal to

    A)  \[\frac{2}{b}\]                 

    B)         \[\frac{3}{b}\]

    C)  \[\frac{b}{3}\]                 

    D)         \[\frac{b}{2}\]

    E)  \[\frac{1}{b}\]

    Correct Answer: A

    Solution :

    Given, \[{{b}^{2}}=ac,x=\frac{a+b}{2}\]and \[y=\frac{b+c}{2}\] \[\therefore \]  \[\frac{1}{x}+\frac{1}{y}=\frac{2}{a+b}+\frac{2}{b+c}\]                 \[=\frac{2(2b+a+c)}{ab+{{b}^{2}}+bc+ac}\]                 \[=\frac{2(2b+a+c)}{ab+2{{b}^{2}}+bc}\]                 \[=\frac{2(2b+a+c)}{b(2b+a+c)}\]                 \[=\frac{2}{b}\]


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