Question

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 $n?$

$1013$

$1014$

$1016$

$2016$

$2025$

Answer: B

Concepts:mean arithmetic sequence bounding to limit cases

Difficulty rating: 1730

Solution:

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 Problem 14

Answer: B

Difficulty rating: 1730

Solution:

Problem 14 in Other Years