Let the sequence `a_1,a_2,a_3, ,a_n` from an A.P. Then the value of `a1 2-a2 2+a3 2-+a2n-1 2-a2n2` is `(2n)/(n-1)(a2n2-a1 2)` (b) `n/(2n-1)(a1 2-a2n2)` `n/(n+1)(a1 2-a2n2)` (d) `n/(n-1)(a1 2+a2n2)`

Options

A. `(n)/(2n+1)(a_(1)^(2)+a_(2n)^(2))`
B. `(2n)/(n+1)(a_(2n)^(2)+a_(1)^(2))`
C. `(n)/(n+1)(a_(1)^(2)+a_(2n)^(2))`
D. none of these

Explanation

To solve the problem, we need to evaluate the expression:

\[ S = a_1^2 - a_2^2 + a_3^2 - a_4^2 + \ldots + a_{2n-1}^2 - a_{2n}^2 \]

Given that the sequence \( a_1, a_2, a_3, \ldots, a_n \) forms an arithmetic progression (A.P.), we can express the terms of the A.P. as follows:

\[ a_k = a_1 + (k-1)d \]

where \( d \) is the common difference.

### Step 1: Express the terms in the sequence

The terms can be expressed as:
- \( a_1 = a_1 \)
- \( a_2 = a_1 + d \)
- \( a_3 = a_1 + 2d \)
- \( a_4 = a_1 + 3d \)
- ...
- \( a_{2n} = a_1 + (2n-1)d \)

### Step 2: Substitute into the expression

Now, substituting these into the expression \( S \):

\[ S = a_1^2 - (a_1 + d)^2 + (a_1 + 2d)^2 - (a_1 + 3d)^2 + \ldots + (a_1 + (2n-2)d)^2 - (a_1 + (2n-1)d)^2 \]

### Step 3: Use the difference of squares formula

Recall the difference of squares formula:

\[ A^2 - B^2 = (A - B)(A + B) \]

Using this, we can rewrite each pair of terms in \( S \):

\[ S = (a_1 - (a_1 + d))(a_1 + (a_1 + d)) + ((a_1 + 2d) - (a_1 + 3d))((a_1 + 2d) + (a_1 + 3d)) + \ldots \]

### Step 4: Simplify each term

Each of these differences simplifies to:

\[ S = -d(2a_1 + d) + (-d)(2a_1 + 5d) + \ldots + (-d)(2a_1 + (2n-1)d) \]

### Step 5: Factor out the common terms

Factoring out \(-d\):

\[ S = -d \left[ (2a_1 + d) + (2a_1 + 5d) + \ldots + (2a_1 + (2n-1)d) \right] \]

### Step 6: Recognize the pattern in the sum

The sum inside the brackets can be expressed as:

\[ S = -d \left[ n(2a_1) + d(1 + 3 + 5 + \ldots + (2n-1)) \right] \]

The sum \( 1 + 3 + 5 + \ldots + (2n-1) \) is the sum of the first \( n \) odd numbers, which equals \( n^2 \).

### Step 7: Substitute back into the expression

Thus, we have:

\[ S = -d \left[ n(2a_1) + d(n^2) \right] \]

### Step 8: Substitute for \( d \)

From the definition of \( d \):

\[ d = \frac{a_{2n} - a_1}{2n - 1} \]

Substituting this back into our expression for \( S \), we can simplify further.

### Final Result

After simplification, we find that:

\[ S = \frac{n}{2n-1}(a_1^2 - a_{2n}^2) \]

Thus, the correct answer is:

\[ \boxed{\frac{n}{2n-1}(a_1^2 - a_{2n}^2)} \]