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

  • question_answer
    The equation \[\left( {{a}^{2}}+{{b}^{2}} \right){{x}^{2}}-2b\left( a+c \right)x+\] \[\left( {{b}^{2}}+{{c}^{2}} \right)=0\] has equal roots. Which one of the following is correct about a, b, and c?

    A) They are in AP

    B) They are in GP

    C) They are in HP

    D) They are neither in AP, nor in GP, nor in HP

    Correct Answer: B

    Solution :

    [b] The given equation \[({{a}^{2}}+{{b}^{2}}){{x}^{2}}-2b(a+c)x+({{b}^{2}}+{{c}^{2}})=0\] has equal roots, so, discriminant = 0 Hence, \[\{2b{{(a+c)}^{2}}-4({{a}^{2}}+{{b}^{2}})({{b}^{2}}+{{c}^{2}})=0\] \[\Rightarrow \,\,\,4{{b}^{2}}({{a}^{2}}+{{c}^{2}}+2ca)-4({{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}+{{b}^{4}}+{{b}^{2}}{{c}^{2}})=0\]\[\Rightarrow \,\,{{b}^{2}}{{a}^{2}}+{{b}^{2}}{{c}^{2}}+2{{b}^{2}}ca-{{a}^{2}}{{b}^{2}}-{{a}^{2}}{{c}^{2}}\]                                    \[-{{b}^{4}}-{{b}^{2}}{{c}^{2}}=0\] \[\Rightarrow \,\,\,\,2{{b}^{2}}ca={{b}^{4}}+{{a}^{2}}{{c}^{2}}\] \[\Rightarrow \,\,\,\,b4-2{{b}^{2}}ca+{{a}^{2}}{{c}^{2}}=0\] \[\Rightarrow \,\,\,\,{{({{b}^{2}}-ac)}^{2}}=0\] \[\Rightarrow \,\,\,\,{{b}^{2}}=ac\] \[\Rightarrow \]  a, b, c are in GP.


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