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Question

Find two numbers whose arithmetic mean is 34 and the geometric mean is 16.

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Solution

Let the two numbers be a and b such that a>b

Then,
a+b2=34 and ab=16

a+b=68.......(1) and ab=256

(ab)2=(a+b)24ab

(ab)2=(68)24×256=3600

ab=60.......(2)

solving eqn (1) & (2), we get,

a=64 and b=4

Hence, required numbers are 64 and 4.

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