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JCECE Engineering JCECE Engineering Solved Paper-2002

  • question_answer
    The number of terms of the \[AP\,\,3,\,\,7,\,\,11,\,\,15...\] to be taken so that the sum is\[210,\,\,\,\text{is}:\]

    A) \[10\]                                   

    B) \[12\]

    C)  -1                                          

    D)  2

    Correct Answer: A

    Solution :

    Given series is   \[3,\,\,7,\,\,11,\,\,15,...\]whose first term is\[3\], common difference \[4\] and sum is \[210\]. Let n be the number of term. \[\therefore \]  \[S=\frac{n}{2}[2a+(n-1)d]\] \[\Rightarrow \]               \[210=\frac{n}{2}[6+(n-1)4]\] \[\Rightarrow \]               \[210=n[3+2n-2]\] \[\Rightarrow \]               \[2{{n}^{2}}+n-210=0\] \[\Rightarrow \]               \[n=10\]


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