Let A, B, C,... X,Y,Z be positive numbers such that AC=B,BD=C...XZ=Y.Given that A+B=1988 find max value of Y+Z?

Options:

A . 994

B . 1988

C . √1988

D . 3976

E . none of these

Answer: Option B : B Solution: Let A = x and C =y Then B = xy D=1x E=1xy F=1y G = x H = xy I = y So we see that this is a sequence with a period six. So A = Y and B = Z Given A+B = 1988, so max value of Y+Z = 1988.

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