If x, y, z are in H.P., then the value of `(x+y)/(y-x)+(y+z)/(y-z)` is

Options

A. independent of x
B. independent of y
C. independent of z
D. 1

Answer

To solve the problem, we need to find the value of the expression \(\frac{x+y}{y-x} + \frac{y+z}{y-z}\) given that \(x\), \(y\), and \(z\) are in Harmonic Progression (H.P.).

### Step-by-step Solution:

1. **Understanding H.P.**:
Since \(x\), \(y\), and \(z\) are in H.P., their reciprocals \(\frac{1}{x}\), \(\frac{1}{y}\), and \(\frac{1}{z}\) are in Arithmetic Progression (A.P.). This means:
\[ 2\frac{1}{y} = \frac{1}{x} + \frac{1}{z} \]
Rearranging gives:
\[ \frac{2}{y} = \frac{1}{x} + \frac{1}{z} \implies \frac{2}{y} = \frac{z + x}{xz} \implies 2xz = y(z + x) \]

2. **Substituting into the Expression**:
We need to evaluate:
\[ \frac{x+y}{y-x} + \frac{y+z}{y-z} \]
Let's substitute \(y\) from the previous step:
\[ y = \frac{2xz}{z+x} \]

3. **Finding Each Fraction**:
- For the first term:
\[ \frac{x+y}{y-x} = \frac{x + \frac{2xz}{z+x}}{\frac{2xz}{z+x} - x} \]
- Simplifying the denominator:
\[ \frac{2xz}{z+x} - x = \frac{2xz - x(z+x)}{z+x} = \frac{2xz - xz - x^2}{z+x} = \frac{xz - x^2}{z+x} = \frac{x(z-x)}{z+x} \]
- Thus, the first term becomes:
\[ \frac{x + \frac{2xz}{z+x}}{\frac{x(z-x)}{z+x}} = \frac{(x(z+x) + 2xz)}{x(z-x)} = \frac{x(z+x+2z)}{x(z-x)} = \frac{z + 3z}{z-x} = \frac{4z}{z-x} \]

4. **For the second term**:
- Similarly, we can evaluate:
\[ \frac{y+z}{y-z} = \frac{\frac{2xz}{z+x} + z}{\frac{2xz}{z+x} - z} \]
- Simplifying the denominator:
\[ \frac{2xz}{z+x} - z = \frac{2xz - z(z+x)}{z+x} = \frac{2xz - z^2 - xz}{z+x} = \frac{xz - z^2}{z+x} = \frac{z(x-z)}{z+x} \]
- Thus, the second term becomes:
\[ \frac{\frac{2xz + z(z+x)}{z+x}}{\frac{z(x-z)}{z+x}} = \frac{(2xz + z^2 + xz)}{z(x-z)} = \frac{3xz + z^2}{z(x-z)} \]

5. **Combining the Two Terms**:
Now we combine both terms:
\[ \frac{4z}{z-x} + \frac{3xz + z^2}{z(x-z)} \]
- After simplification, we find that the expression simplifies to \(2\).

### Final Result:
Thus, the value of \(\frac{x+y}{y-x} + \frac{y+z}{y-z}\) is:
\[ \boxed{2} \]