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JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Evening)

  • question_answer
    Let a, b and c be the\[{{7}^{th}}\], \[{{11}^{th}}\] and \[{{13}^{th}}\] terms respectively of a non-constant A.P. If these are also the three consecutive terms of a G.P., then \[\frac{a}{c}\] is equal to: [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

    A) \[\frac{7}{13}\]                                     

    B)               2

    C)               4                     

    D)               \[\frac{1}{2}\]

    Correct Answer: C

    Solution :

    \[{{T}_{7}}=A+6d=a;{{T}_{11}}=A+10d=b;\] \[{{T}_{13}}=A+12d=c\] Now a, b, c are in G.P. \[\therefore \,\,\,\,\text{ }{{b}^{2}}=ac\] \[\Rightarrow \,\,\,\,{{\left( A+10d \right)}^{2}}=\left( A+6d \right)\left( A+12d \right)\] \[\Rightarrow \,\,\,\,{{A}^{2}}+100{{d}^{2}}+20Ad={{A}^{2}}+18Ad+72{{d}^{2}}\] \[\Rightarrow \,\,\,\,\,A+14d=0,\text{ }A=-14d\] \[\frac{a}{c}\,=\,\,\frac{A+6d}{A+12d}\,=\,\frac{-8d}{-2d}\,\,=\,\,4\]    


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