Question

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 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 $n?$

$1013$

$1014$

$1016$

$2016$

$2025$

Solution:

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 $x_{1014} = 1.$ We can pick the first three terms as decreasing integers above $2021$ summing to $6075,$ so $n = 1014$ is reachable. Thus, B is the correct answer.

2025 AMC 10B Problem 17

Solution:

Problem 17 in Other Years