Question
If a, b, c are three distinct positive real numbers which are in H.P., then 3a+2b2a−b+3c+2b2c−b is
Answer: Option D
:
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+2b2a−b+3c+2b2c−b then it reduces to
10+14q2p2−q2 which is obviously greater than
10(as p > q > 0).
Trick: Put a = 1, b = 12, c = 13.
The expression has the value 3+12−12+1+123−12 = 83+12>0
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:
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+2b2a−b+3c+2b2c−b then it reduces to
10+14q2p2−q2 which is obviously greater than
10(as p > q > 0).
Trick: Put a = 1, b = 12, c = 13.
The expression has the value 3+12−12+1+123−12 = 83+12>0
Was this answer helpful ?
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