If a,b,c are three distinct positive real numbers which are in H.P., then `(3a + 2b)/(2a - b) + (3c + 2b)/(2c - b)` is

Options

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

Answer

To solve the problem, we need to analyze the expression given that \( a, b, c \) are three distinct positive real numbers in Harmonic Progression (H.P.).

### Step-by-step Solution:

1. **Understanding H.P.**:
If \( a, b, c \) are in H.P., then the reciprocals \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in Arithmetic Progression (A.P.). This means:
\[ \frac{1}{b} = \frac{\frac{1}{a} + \frac{1}{c}}{2} \implies b = \frac{2ac}{a+c} \]

2. **Substituting \( b \)**:
We substitute \( b \) in the expression:
\[ \frac{3a + 2b}{2a - b} + \frac{3c + 2b}{2c - b} \]
becomes:
\[ \frac{3a + 2\left(\frac{2ac}{a+c}\right)}{2a - \frac{2ac}{a+c}} + \frac{3c + 2\left(\frac{2ac}{a+c}\right)}{2c - \frac{2ac}{a+c}} \]

3. **Simplifying the First Fraction**:
The first term simplifies as follows:
\[ \frac{3a + \frac{4ac}{a+c}}{2a - \frac{2ac}{a+c}} = \frac{(3a(a+c) + 4ac)}{(2a(a+c) - 2ac)} = \frac{3a^2 + 3ac + 4ac}{2a^2 + 2ac - 2ac} = \frac{3a^2 + 7ac}{2a^2} \]

4. **Simplifying the Second Fraction**:
The second term simplifies similarly:
\[ \frac{3c + \frac{4ac}{a+c}}{2c - \frac{2ac}{a+c}} = \frac{(3c(a+c) + 4ac)}{(2c(a+c) - 2ac)} = \frac{3c^2 + 7ac}{2c^2} \]

5. **Combining the Two Fractions**:
Now we combine both fractions:
\[ \frac{3a^2 + 7ac}{2a^2} + \frac{3c^2 + 7ac}{2c^2} \]
Finding a common denominator gives:
\[ \frac{(3a^2 + 7ac)c^2 + (3c^2 + 7ac)a^2}{2a^2c^2} \]

6. **Expanding the Numerator**:
Expanding the numerator:
\[ 3a^2c^2 + 7ac^3 + 3c^2a^2 + 7a^2c = 6a^2c^2 + 7ac(a + c) \]

7. **Final Expression**:
Thus, we have:
\[ \frac{6a^2c^2 + 7ac(a+c)}{2a^2c^2} \]

8. **Conclusion**:
The final expression simplifies to:
\[ 3 + \frac{7(a+c)}{2ac} \]

### Final Result:
The expression evaluates to:
\[ \frac{(3a + 2b)}{(2a - b)} + \frac{(3c + 2b)}{(2c - b)} = 3 + \frac{7(a+c)}{2ac} \]