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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2013

  • question_answer
        For an increasing AP\[{{a}_{1}},{{a}_{2}},.....,{{a}_{n}},\]if\[{{a}_{1}}+{{a}_{2}}+{{a}_{3}}=-12\]and\[{{a}_{1}}{{a}_{3}}{{a}_{5}}=80,\]then which of the following does not hold?

    A)  \[{{a}_{1}}=-10\]            

    B)  \[{{a}_{2}}=-1\]

    C)  \[{{a}_{3}}=-4\]                               

    D)  \[{{a}_{5}}=2\]

    Correct Answer: B

    Solution :

                    \[\because \]\[{{a}_{1}}+{{a}_{3}}+{{a}_{5}}=-12\] \[a+(a+2d)+(a+4d)=-12\]                  \[(\because d>0)\] \[\Rightarrow \]          \[a+2d=-4\]                               ...(i) and         \[{{a}_{1}}{{a}_{3}}{{a}_{5}}=80\] \[\Rightarrow \]\[a(a+2d)(a+4d)=80\] \[\Rightarrow \]               \[a(-4)(-4-2d+4d)=80\] [from Eq. (i)] \[\Rightarrow \]               \[(-4-2d)(-4)(-4-2d+4d)=80\] \[\Rightarrow \]               \[4(4+2d)(-4+2d)=80\] \[\Rightarrow \]               \[4(4{{d}^{2}}-16)=80\] \[\Rightarrow \]               \[4{{d}^{2}}=16+20=36\] \[\Rightarrow \]               \[{{d}^{2}}=9\] \[\therefore \]  \[d=\pm 3\] Since, AP is increasing, then \[d=+3;a=-10\] \[\therefore \]  \[{{a}_{1}}=-10;{{a}_{2}}=-7\]                 \[{{a}_{3}}=a+2d=-10+6=-4\]                 \[{{a}_{5}}=a+4d=-10+8=-2\]


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