Question

考虑一个由 $n$ 个正整数组成的递减序列 $x_1 \gt x_2 \gt x_3 \gt \cdots \gt x_n$ ，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 $4 \le k \le n$ ，序列前 $k$ 项的平均数小 $1$ ，这是相对于前 $k - 1$ 项的平均数而言。

$n$ 的最大可能值是多少？

Consider a decreasing sequence of $n$ positive integers $x_1 \gt x_2 \gt x_3 \gt \cdots \gt x_n$ that satisfies the following 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：平均数求和极限情形界定

难度评级：1910

解答：

令 $A_k$ 为前 $k$ 项平均数。已知 $A_3 = 2025$ ，且 $A_k = A_{k-1} - 1$ （当 $k \ge 4$ ），所以 $A_k = 2028 - k$ 。前 k 项和为 $S_k = k(2028 - k)$ ，因此当 $k \ge 4$ 时， $x_k = S_k - S_{k-1} = 2029 - 2k$ ，也就是 $x_4 = 2021, x_5 = 2019, \ldots$ 。这些项保持为正当且仅当 $2029 - 2k \gt 0$ ，即 $k \le 1014$ ，且 $x_{1014} = 1$ 。可以选择前三项为大于 $(x_1,x_2,x_3) = (2030,2023,2022)$ 的递减整数并使它们和为 $6075$ ，所以 $n = 1014$ 可以达到。因此正确答案是 B。 $x_4 = 2021$ 。

Let $A_k$ be the average of the first $k$ terms. Then $A_3 = 2025$ and $A_k = A_{k-1} - 1$ for $k \ge 4,$ so $A_k = 2028 - k.$ The partial sum is $S_k = k(2028 - k),$ and for $k \ge 4$ the terms are $x_k = S_k - S_{k-1} = 2029 - 2k,$ namely $x_4 = 2021, x_5 = 2019, \ldots.$ These stay positive as long as $2029 - 2k \gt 0,$ that is $k \le 1014,$ with $x_{1014} = 1.$ This bound is attainable: take $(x_1,x_2,x_3) = (2030,2023,2022),$ which is decreasing, sums to $6075,$ and lies above $x_4 = 2021.$ Thus the greatest possible value is $n = 1014,$ so B is the correct answer.

2025 AMC 10B 第 17 题

答案：B

难度评级：1910

解答：

其他年份的第 17 题