🔥 New! 2025 AMC 10 B

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Problem 17 Easy

Consider a decreasing sequence of n positive integerss x _ { 1 } > x _ { 2 } > x _ { 3 } > \cdots > 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 \leqslant k \leqslant 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?

  • A.

    1013

  • B.

    1014

  • C.

    1016

  • D.

    2016

  • E.

    2025

Answer:B
Solution

Let a_k be the average of the first k terms.

Given a_3 = 2025 and a_k = a_{k-1} - 1 for k \geq 4.

Thus a_k = 2025 - (k-3) = 2028 - k for k \geq 3.

x_k = k \cdot a_k - (k-1) \cdot a_{k-1} = a_k + (k-1) \cdot (a_k - a_{k-1})

= 2028 - k - k + 1 = 2029 - 2k for k \geq 4

For x_k > 0, we need 2029 - 2k > 0, which gives k \leq 1014.

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