Question

If a, b, c are three distinct positive real numbers which are in H.P., then $\frac{3a+2b}{2a-b}+\frac{3c+2b}{2c-b}$ is

• A. Greater than or equal to 10
• B. Less than or equal to 10
• C. Only equal to 10
• D. None of these

Answer 1

The correct option is D

None of these

we have $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P. Let $\frac{1}{a}$ = p - q, $\frac{1}{b}$ = p and

$\frac{1}{c}$ = p + q, where p,q > 0 and p > q. Now, substitute these

values in $\frac{3a+2b}{2a-b}+\frac{3c+2b}{2c-b}$ then it reduces to

$10+\frac{14q^2}{p^2-q^2}$ which is obviously greater than

10(as p > q > 0).

Trick: Put a = 1, b = $\frac{1}{2}$, c = $\frac{1}{3}$.

The expression has the value $\frac{3+1}{2-\frac{1}{2}}+\frac{1+1}{\frac{2}{3}-\frac{1}{2}}$ = $\frac{8}{3}+12>0$