2025 AMC 12B Problem 14

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

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.

Problem 14 in other years