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Let the sums received by P, Q, R, S, T and U are a+5d, a+4d, a+3d, a+2d, a+d and a respectively a+5d + a+4d + a+3d + a+2d + a+d + a = 558 6a+15d = 558 0r 2a+5d = 186 And also, (a+5d) – (a+2d) = 36 3d = 36 d = 12 now from 2a+5d = 186 we get a = 63 sum of Q = a+4d sum of U = a sum of Q and U = a+a+4d = 174
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