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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 30

  • question_answer
    If the first, second and last terms of an A.P. be a, b, 2a respectively, then its sum will be

    A) \[\frac{ab}{b-a}\]          

    B) \[\frac{ab}{2(b-a)}\]

    C) \[\frac{3ab}{2(b-a)}\]    

    D) \[\frac{3ab}{4(b-a)}\]

    Correct Answer: C

    Solution :

    [c] : We have first term = a                      ...(i) Second term = b                                    ...(ii) and last term = 2a                                  ...(iii) Let d be the common difference. From (i), (ii) and (iii), d = (b - a) and \[n=\frac{b}{b-a}\]. Then, sum \[(S)=\frac{n}{2}[a+l]\] \[=\frac{b}{2(b-a)}[a+2a]=\frac{3ab}{2(b-a)}\]


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