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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    The first four terms of an AP are \[a,9,3a-b,3a\]\[+b\]. The 2011th term of the AP is

    A)  2015                                     

    B)  4025

    C)  5030                     

    D)         6035

    E)  8045

    Correct Answer: E

    Solution :

    Given first four term of an AP \[a,\text{ }9,\text{ }3a-b,\text{ }3a+b\] \[\Rightarrow \]               \[2-9=a+3a-b,4a-b=18\]       ...(i) and  \[2(3a-b)=9+3a+b\] \[\Rightarrow \]    \[6a-2b=9+3a+b,3a-3b=9\] \[\Rightarrow \]           \[a-b=3\]                                ...(ii) On solving Eqs. (i) and (ii), \[3a=15\Rightarrow a=5\,and\,b=2\] The series becomes, 5, 9, 13, 17. First term = 5 Common difference = 4 Now, \[{{T}_{2011}}=5(2011+1).4\] \[=5+8040=8045\]


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