In an arithmetical progression `a_(1), a_(2), a_(3),…, S = a_(1)^(2) - a_(2)^(2) + a_(3)^(2) - a_(4)^(2) + …. - a_(2k)^(2) =`

Options

A. `(k)/(2 k - 1) (a_(1)^(2) - a_(2k)^(2))`
B. `(2k)/(k - 1) (a_(2k)^(2) - a_(1)^(2))`
C. `(k)/(k + 1) (a_(1)^(2) + a_(2k)^(2))`
D. none

Answer

To solve the problem, we need to evaluate the sum \( S = a_1^2 - a_2^2 + a_3^2 - a_4^2 + \ldots - a_{2k}^2 \) for an arithmetic progression (AP).

### Step-by-Step Solution:

1. **Identify the terms of the AP**:
In an arithmetic progression, the \( n \)-th term can be expressed as:
\[ a_n = a_1 + (n-1)d \]
where \( a_1 \) is the first term and \( d \) is the common difference.

2. **Express the squares of the terms**:
We can express the squares of the terms in the sum:
\[ a_1^2, a_2^2, a_3^2, \ldots, a_{2k}^2 \]
Specifically:
\[ a_2 = a_1 + d, \quad a_3 = a_1 + 2d, \quad a_4 = a_1 + 3d, \ldots, a_{2k} = a_1 + (2k-1)d \]

3. **Substitute the terms into the sum**:
The sum \( S \) can be rewritten as:
\[ S = a_1^2 - (a_1 + d)^2 + (a_1 + 2d)^2 - (a_1 + 3d)^2 + \ldots - (a_1 + (2k-1)d)^2 \]

4. **Expand the squares**:
Expanding each term gives:
\[ S = a_1^2 - (a_1^2 + 2a_1d + d^2) + (a_1^2 + 4a_1d + 4d^2) - (a_1^2 + 6a_1d + 9d^2) + \ldots \]

5. **Group the terms**:
Notice that the \( a_1^2 \) terms will cancel out in pairs. We can group the terms based on their coefficients:
\[ S = (a_1^2 - a_1^2) + (4a_1d - 2a_1d) + (0 - 6a_1d) + \ldots - (2k-1)^2d^2 \]

6. **Factor out common terms**:
Factor out \( -d \) from the entire expression:
\[ S = -d \left( \text{sum of coefficients} \right) \]

7. **Calculate the sum of coefficients**:
The coefficients can be calculated as:
\[ S = -d \cdot \left( \text{sum of first } 2k \text{ terms with alternating signs} \right) \]

8. **Final expression**:
After calculating the sum, we find:
\[ S = -kd(a_1 + a_{2k}) \]
where \( a_{2k} = a_1 + (2k-1)d \).

### Conclusion:
Thus, the final expression for \( S \) is:
\[ S = -kd(a_1 + a_1 + (2k-1)d) = -kd(2a_1 + (2k-1)d) \]