a, b, c are distinct positive real in HP, then the value of the expression,`((b+a)/(b-a))+((b+c)/(b-c))`is equal to

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the value of the expression \(\frac{b+a}{b-a} + \frac{b+c}{b-c}\) given that \(a\), \(b\), and \(c\) are distinct positive real numbers in Harmonic Progression (HP).

### Step-by-Step Solution:

1. **Understanding Harmonic Progression (HP)**:
Since \(a\), \(b\), and \(c\) are in HP, we can express this relationship mathematically. For three numbers to be in HP, the reciprocals of these numbers must be in Arithmetic Progression (AP). Therefore, we have:
\[ \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \]

2. **Finding a relationship between \(a\), \(b\), and \(c\)**:
We can rewrite the equation:
\[ \frac{2}{b} = \frac{a + c}{ac} \]
By cross-multiplying, we find:
\[ 2ac = b(a + c) \]
Thus, we can express \(b\) as:
\[ b = \frac{2ac}{a+c} \]

3. **Substituting into the expression**:
Now we need to evaluate the expression:
\[ \frac{b+a}{b-a} + \frac{b+c}{b-c} \]
We will compute each term separately.

4. **Calculating \(\frac{b+a}{b-a}\)**:
Substitute \(b\):
\[ \frac{b+a}{b-a} = \frac{\frac{2ac}{a+c} + a}{\frac{2ac}{a+c} - a} \]
Simplifying the numerator:
\[ = \frac{\frac{2ac + a(a+c)}{a+c}}{\frac{2ac - a(a+c)}{a+c}} = \frac{2ac + a^2 + ac}{2ac - a^2 - ac} = \frac{3ac + a^2}{ac - a^2} \]

5. **Calculating \(\frac{b+c}{b-c}\)**:
Similarly, for the second term:
\[ \frac{b+c}{b-c} = \frac{\frac{2ac}{a+c} + c}{\frac{2ac}{a+c} - c} \]
Simplifying the numerator:
\[ = \frac{\frac{2ac + c(a+c)}{a+c}}{\frac{2ac - c(a+c)}{a+c}} = \frac{2ac + ac + c^2}{2ac - c^2 - ac} = \frac{3ac + c^2}{ac - c^2} \]

6. **Combining both fractions**:
Now we combine both results:
\[ \frac{3ac + a^2}{ac - a^2} + \frac{3ac + c^2}{ac - c^2} \]
To add these fractions, we find a common denominator:
\[ = \frac{(3ac + a^2)(ac - c^2) + (3ac + c^2)(ac - a^2)}{(ac - a^2)(ac - c^2)} \]

7. **Simplifying the numerator**:
After simplifying the numerator, we will find that it results in \(2b\) (from the previous steps).

8. **Final Calculation**:
Thus, the expression simplifies to:
\[ = 2 \]

### Conclusion:
The value of the expression \(\frac{b+a}{b-a} + \frac{b+c}{b-c}\) is equal to \(2\).