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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2002

  • question_answer
    The first three terms in the expansion of\[{{(1+ax)}^{n}}(n\ne 0)\]are\[1,6x\]and\[16{{x}^{2}}\].Then the values of a and n are respectively:

    A)  2 and 9                                

    B)                         3 and 2

    C)                         \[\frac{2}{3}\]and 9        

    D)                        \[\frac{3}{2}\]and 6

    E)                         \[\frac{-2}{3}\] and 9

    Correct Answer: C

    Solution :

                    \[\because \] \[{{(1+ax)}^{n}}=1+axn+\frac{n(n-1)}{2!}{{(ax)}^{2}}+....\] Given that \[a\,x\,n=6x\] \[\Rightarrow \]         \[an=6\]                                         ...(i) and       \[\frac{{{a}^{2}}n(n-1)}{2}=16\]                      ?..(ii) On solving Eqs. (i) and (ii), we get \[a=\frac{2}{3}\]and\[n=9\]


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