Question

考虑一个由 $n$ 个正整数组成的递减序列 $x_1 \gt x_2 \gt \cdots \gt x_n$ ，满足以下两个条件：

• 序列前 $3$ 项的平均数（算术平均值）是 $2025$ 。

• 对所有满足 $4 \le k \le n$ 的整数，序列前 $k$ 项的平均数比前 $k-1$ 项的平均数小 $1$ 。

$n$ 的最大可能值是多少？

Consider a decreasing sequence of $n$ positive integers $x_1 \gt x_2 \gt \cdots \gt x_n$ that satisfies the following two conditions:

• The average (arithmetic mean) of the first $3$ terms in the sequence is $2025.$

• For all $4 \le k \le n,$ the average of the first $k$ terms in the sequence is $1$ less than the average of the first $k-1$ terms in the sequence.

What is the greatest possible value of $n?$

$1013$

$1014$

$1016$

$2016$

$2025$

答案：B

知识点：平均数等差数列极限情形界定

难度评级：1730

解答：

对 $k \ge 3$ 前 $k$ 项的平均数为 $A_k = 2028 - k$ ，所以部分和为 $S_k = k(2028 - k)$ 对 $k \ge 4$ $x_k = S_k - S_{k-1} = 2029 - 2k$ 它为正数当且仅当 $k \le 1014$ 例如取 $x_1, x_2, x_3 = 2030, 2023, 2022$ ，可使整个序列严格递减，所以 $n$ 的最大可能值为 $1014$ 。

所以正确答案是 B。

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.

2025 AMC 12B 第 14 题

答案：B

难度评级：1730

解答：

其他年份的第 14 题