If `x^(2),y^(2),z^(2)` are in AP, then `y+z,z+x,x+y` are in

Options

A. AP
B. HP
C. GP
D. None of these

Explanation

To solve the problem, we need to determine the relationship between the expressions \(y + z\), \(z + x\), and \(x + y\) given that \(x^2\), \(y^2\), and \(z^2\) are in arithmetic progression (AP).

### Step-by-Step Solution:

1. **Understanding the Condition of AP**:
Since \(x^2\), \(y^2\), and \(z^2\) are in AP, we can write the condition for AP:
\[ 2y^2 = x^2 + z^2 \]

2. **Expressing the Terms**:
We need to analyze the expressions \(y + z\), \(z + x\), and \(x + y\). Let's denote:
- \(A = y + z\)
- \(B = z + x\)
- \(C = x + y\)

3. **Finding the Mean**:
To check if \(A\), \(B\), and \(C\) are in AP, we need to check if:
\[ 2B = A + C \]
Substituting the expressions for \(A\), \(B\), and \(C\):
\[ 2(z + x) = (y + z) + (x + y) \]

4. **Simplifying the Equation**:
Expanding both sides:
\[ 2z + 2x = y + z + x + y \]
This simplifies to:
\[ 2z + 2x = 2y + z + x \]

5. **Rearranging the Terms**:
Rearranging gives:
\[ 2z + 2x - z - x = 2y \]
Simplifying further:
\[ z + x = 2y \]

6. **Using the AP Condition**:
From the earlier step, we know:
\[ 2y^2 = x^2 + z^2 \]
This implies that:
\[ y = \frac{x + z}{2} \]
Therefore, substituting \(y\) back into the equation \(z + x = 2y\) confirms:
\[ z + x = 2\left(\frac{x + z}{2}\right) \]
This holds true.

7. **Conclusion**:
Since we have shown that \(2(z + x) = (y + z) + (x + y)\), we conclude that:
\[ y + z, z + x, x + y \text{ are in AP.} \]

### Final Answer:
The expressions \(y + z\), \(z + x\), and \(x + y\) are in **Arithmetic Progression (AP)**.