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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