If `x=(1)/(1^(2))+(1)/(3^(2))+(1)/(5^(2))+....` , `y=(1)/(1^(2))+(3)/(2^(2))+(1)/(3^(2))+(3)/(4^(2))+....` and `z=(1)/(1^(2))-(1)/(2^(2))+(1)/(3^(2))-(1)/(4^(2))+...` then

Options

A. `x`, `y`,`z` are in `A.P.`
B. `(y)/(6)`, `(x)/(3)`,`(z)/(2)` are in `A.P.`
C. `(y)/(6)`, `(x)/(3)`,`(z)/(2)` are in `A.P.`
D. `6y`, `3x`, `2z` are in `H.P.`

Explanation

To solve the problem, we need to analyze the series for \( x \), \( y \), and \( z \) and determine their relationships.

### Step 1: Define the series

1. **Define \( x \)**:
\[ x = \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots \]

2. **Define \( y \)**:
\[ y = \sum_{n=1}^{\infty} \frac{1}{n^2} + \sum_{n=1}^{\infty} \frac{3}{(n+1)^2} = \frac{1}{1^2} + \frac{3}{2^2} + \frac{1}{3^2} + \frac{3}{4^2} + \ldots \]

3. **Define \( z \)**:
\[ z = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2} = \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \ldots \]

### Step 2: Analyze \( y - x \)

To find the relationship between \( y \) and \( x \):
\[ y - x = \left( \frac{1}{1^2} + \frac{3}{2^2} + \frac{1}{3^2} + \frac{3}{4^2} + \ldots \right) - \left( \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots \right) \]

Notice that the terms with odd denominators in \( x \) will cancel out with those in \( y \). The remaining terms will be:
\[ y - x = \sum_{n=1}^{\infty} \frac{3}{(2n)^2} = 3 \sum_{n=1}^{\infty} \frac{1}{(2n)^2} = 3 \cdot \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{3}{4} \cdot \frac{\pi^2}{6} \]

### Step 3: Analyze \( x - z \)

Now, consider \( x - z \):
\[ x - z = \left( \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots \right) - \left( \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \ldots \right) \]

Here, the terms with \( \frac{1}{1^2} \) and \( \frac{1}{3^2} \) will cancel out, and we are left with:
\[ x - z = \sum_{n=1}^{\infty} \frac{1}{(2n)^2} = \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{24} \]

### Step 4: Establish the relationship

From the previous steps, we have:
1. \( y - x = \frac{3}{4} \cdot \frac{\pi^2}{6} \)
2. \( x - z = \frac{\pi^2}{24} \)

Now, we can express these relationships:
\[ y - x = 3(x - z) \]

This implies:
\[ y + 3z = 4x \]

### Step 5: Conclusion

From the equation \( y + 3z = 4x \), we can rearrange it to show that:
\[ \frac{y}{6} + \frac{3z}{6} = \frac{4x}{6} \]

This indicates that \( \frac{y}{6}, \frac{3z}{6}, \frac{4x}{6} \) are in Arithmetic Progression (AP).

Thus, the correct answer is that \( \frac{y}{6}, \frac{3z}{6}, \frac{4x}{6} \) are in AP.