The sequence a_1, a_2, a_3, … satisfies a_1 = 19, a_9 = 99, and, for all n ≥ 3, a_n is the arithmetic mean of the first n - 1 terms. Find a_2.

1999 AMC 12 Problem 20 — Solution

1999 AMC 12 Problem 20

Below is the professionally curated solution for Problem 20 of the 1999 AMC 12, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1999 AMC 12 solutions, or check the answer key.

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Concepts:recursion mean invariant

Difficulty rating: 1740

20.

The sequence $a_1, a_2, a_3, \ldots$ satisfies $a_1 = 19,$ $a_9 = 99,$ and, for all $n \ge 3,$ $a_n$ is the arithmetic mean of the first $n - 1$ terms. Find $a_2.$

$29$

$59$

$79$

$99$

$179$

Solution:

For $n \ge 3,$ $(n - 1)a_n = a_1 + \cdots + a_{n-1}.$ Then $a_{n+1} = \dfrac{(n-1)a_n + a_n}{n} = a_n,$ so the sequence is constant from $a_3$ onward. Hence $a_3 = a_9 = 99.$

Since $a_3 = \dfrac{a_1 + a_2}{2} = \dfrac{19 + a_2}{2} = 99,$ we get $a_2 = 179.$

Thus, the correct answer is E.

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