2025 AMC 12B 第 14 题

Let $A_k$ be the average of the first $k$ terms; the second condition gives $A_k = 2028 - k$ for $k \ge 3$

From $S_k = k A_k$ recover $x_k = 2029 - 2k$ for $k \ge 4,$ then require every term to stay positive

The average of the first $k$ terms is $A_k = 2028 - k$ for $k \ge 3,$ so the partial sum is $S_k = k(2028 - k).$ For $k \ge 4,$ $x_k = S_k - S_{k-1} = 2029 - 2k,$ which is positive exactly when $k \le 1014.$ A valid start such as $x_1, x_2, x_3 = 2030, 2023, 2022$ keeps the whole sequence strictly decreasing, so the greatest possible $n$ is $1014.$

Thus, the correct answer is B.