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JEE Main & Advanced JEE Main Solved Paper-2017

  • question_answer
    For any three positive real numbers a, b and c, \[9(25{{a}^{2}}+{{b}^{2}})+25({{c}^{2}}-3ac)=15b(3a+c).\] Then:                              JEE Main Solved Paper-2017

    A)  a, b and c are in G.P.     

    B)  b, c and a are in G.P.

    C)  b, c and a are in A.P.

    D)  a, b and c are in A.P.

    Correct Answer: C

    Solution :

     \[{{(15a)}^{2}}+{{(3b)}^{2}}+{{(5c)}^{2}}-(15a)(5c)-(15a)(3b)\]                 \[-(3b)(5c)=0\]  \[\frac{1}{2}[{{(15a-3b)}^{2}}+{{(3b-5c)}^{2}}+{{(5c-15a)}^{2}}]=0\] it is possible when 15a = 3b = 5c \[\therefore \]  \[b=\frac{5c}{3},\,a=\frac{c}{3}\]                 \[a+b=2c\] \[\Rightarrow \]               \[b,c,\,a\]in \[A.P.\]


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