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.You need to login to perform this action.
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