404
442
455
415
Arithmetic Progression (AP): 3, 7, 11, ...
First term: \( a = 3 \)
Common difference: \( d = 4 \)
The general formula for the sum of the first \( n \) terms of an AP is \( S_n = \frac{n}{2} \left(2a + (n - 1)d \right).
Substituting \( a = 3 \) and \( d = 4 \) yields \( S_n = \frac{n}{2} \left(2 \times 3 + (n - 1) \times 4 \right) = \frac{n}{2}(6 + 4n - 4) = \frac{n}{2}(4n + 2) = n(2n + 1) \).
We need to find \( \frac{1}{25} \sum_{n=1}^{25} A_n \), where \( A_n = S_n = n(2n + 1) \).
This requires calculating \( \sum_{n=1}^{25} n(2n + 1) \).
Expanding the term gives \( \sum_{n=1}^{25} (2n^2 + n) \).
This can be split into two separate summations: \( \sum_{n=1}^{25} 2n^2 + \sum_{n=1}^{25} n = 2 \sum_{n=1}^{25} n^2 + \sum_{n=1}^{25} n \).
The sum of the squares of the first \( n \) natural numbers is \( \sum_{n=1}^{25} n^2 = \frac{25 \cdot 26 \cdot 51}{6} = 5525 \).
Therefore, \( 2 \sum_{n=1}^{25} n^2 = 2 \times 5525 = 11050 >.
The sum of the first \( n \) natural numbers is \( \sum_{n=1}^{25} n = \frac{25 \cdot 26}{2} = 325 >.
Adding these two results gives \( \sum_{n=1}^{25} n(2n + 1) = 11050 + 325 = 11375 >.
The required expression is \( \frac{1}{25} \sum_{n=1}^{25} A_n = \frac{11375}{25} = \boxed{455} >.
The final answer is 455.