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Question

If a, b, c are three distinvt positive real numbers which are in H.P., then 3a+2b2ab+3c+2b2cb is


A

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B

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C

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D

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Solution

The correct option is D


we have 1a,1b,1c are in A.P. Let 1a = p - q, 1b = p and

1c = p + q, where p,q > 0 and p > q. Now, substitute these

values in 3a+2b2ab+3c+2b2cb then it reduces to

10+14q2p2q2 which is obviously greater than

10(as p > q > 0).

Trick: Put a = 1, b = 12, c = 13.

The expression has the value 3+1212+1+12312 = 83+12>0


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