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11th Class Mathematics Sequence and Series Question Bank Sequence and Series (A.P., G.P. and H.P.)

  • question_answer
    Divide 124 into four parts in such way that they are in AP and the product of the first and the 4th part is 128 less than the product of the 2nd and the 3rd parts, the parts are:

    A)  27, 35, 45, 17              

    B)  19, 27, 35, 43  

    C)  21, 29, 33, 41              

    D)  13, 33, 37, 41

    Correct Answer: B

    Solution :

    (b): Let the four parts be \[\left( a-3d \right),\left( a-d \right),\left( a+d \right)\] and\[\left( a+3d \right)\]. The sum of these four parts is 124. (Here the mathematical ingenuity is to select AP as shown above) so that their sum simply becomes 4a. Note that the common difference is 2d). \[\therefore 4a=124\Rightarrow a=31\]; Also, \[(a-3d)(a+3d)=(a-d)(a+d)-128\] \[\Rightarrow \]\[{{a}^{2}}-9{{d}^{2}}={{a}^{2}}-{{d}^{2}}-128\] \[\Rightarrow 8{{d}^{2}}=128\Rightarrow d=\pm 4\]. As a = 31, taking d = 4, then four parts are 19, 27, 35 and 43.   


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