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 given for \( x \), \( y \), and \( z \) and find relationships between them.

### Step 1: Define the series
We have:
- \( x = \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots \) (Sum of reciprocals of squares of odd numbers)
- \( y = \frac{1}{1^2} + \frac{3}{2^2} + \frac{1}{3^2} + \frac{3}{4^2} + \ldots \) (Sum of reciprocals of squares of odd and even numbers with alternating coefficients)
- \( z = \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \ldots \) (Alternating series of reciprocals of squares)

### Step 2: Analyze \( x - y \)
To find \( x - y \):
- The odd terms in \( y \) will cancel out with the corresponding terms in \( x \).
- The remaining terms in \( y \) will be the even terms multiplied by 3:
\[ x - y = -3 \left( \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \ldots \right) \]
This series can be expressed as:
\[ x - y = -3 \cdot \frac{1}{4} \left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots \right) = -\frac{3}{4} \cdot \frac{\pi^2}{6} \]

### Step 3: Analyze \( x - z \)
Now, for \( x - z \):
- The odd terms in \( z \) will also cancel out with the corresponding terms in \( x \).
- The remaining terms in \( z \) will be the even terms:
\[ x - z = \left( \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \ldots \right) \]
This series can be expressed as:
\[ x - z = \frac{1}{4} \left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots \right) = \frac{1}{4} \cdot \frac{\pi^2}{6} \]

### Step 4: Relate \( x - y \) and \( x - z \)
From the results of \( x - y \) and \( x - z \):
\[ x - y = -3(x - z) \]
Substituting the expressions we derived:
\[ -\frac{3}{4} \cdot \frac{\pi^2}{6} = -3 \cdot \frac{1}{4} \cdot \frac{\pi^2}{6} \]
This shows a consistent relationship between the three series.

### Step 5: Final relationship
From the derived relationships:
\[ x - y = -3(x - z) \]
We can express this as:
\[ x - y + 3(x - z) = 0 \]
This leads us to conclude:
\[ 4x = y + 3z \]
Dividing through by 6 gives:
\[ \frac{2x}{3} = \frac{y}{6} + \frac{z}{2} \]

### Conclusion
The relationship \( \frac{y}{6} = \frac{2x}{3} - \frac{z}{2} \) indicates that \( y/6 \), \( x/3 \), and \( z/2 \) are in arithmetic progression.