2025 AMC 10B Problem 17

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 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?

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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,

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.

Tagged: mean · summation · bounding to limit cases