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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2002

  • question_answer
    The number of terms of the AP series 3, 7, 11, 15, ... to be taken so that the sum is 406, is:

    A)  5                                            

    B)                         10

    C)                         12                                         

    D)                         14

    E)                         20

    Correct Answer: D

    Solution :

    Given that, \[a=3,\text{ }d=4,\text{ }5=406\] \[\therefore \] \[406=\frac{n}{2}[6+(n-1)4]\] \[\Rightarrow \]               \[406=n(2n+1)\] \[\Rightarrow \]               \[2{{n}^{2}}+n-406=0\] \[\Rightarrow \]               \[(n-14)(2n+29)=0\] \[\Rightarrow \]               \[n=14\]                                               \[\left( \because n\ne -\frac{29}{2} \right)\]


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