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JEE Main & Advanced Mathematics Sequence & Series Question Bank Arithmetic Progression

  • question_answer
    Three numbers are in A.P. whose sum is 33 and product is  792, then the smallest number from these numbers is [RPET 1988]

    A) 4

    B) 8

    C) 11

    D) 14

    Correct Answer: A

    Solution :

    Suppose that three numbers are \[a+d,\ a,\ a-d,\] therefore \[a+d+a+a-d=33\]\[\Rightarrow \]\[a=11\] \[a(a+d)(a-d)=792\]\[\Rightarrow \]\[11(121-{{d}^{2}})=792\]\[\Rightarrow \]\[d=7\] Then required numbers are 4, 11, 18. Hence smallest number is 4.


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