2004 AIME II Problems/Problem 9

Problem

A sequence of positive integers with $a_1=1$ and $a_9+a_{10}=646$ is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all $n\ge1,$ the terms $a_{2n-1}, a_{2n}, a_{2n+1}$ are in geometric progression, and the terms $a_{2n}, a_{2n+1},$ and $a_{2n+2}$ are in arithmetic progression. Let $a_n$ be the greatest term in this sequence that is less than $1000$ . Find $n+a_n.$

Solution

Let $x = a_2$ ; then solving for the next several terms, we find that $a_3 = x^2,\ a_4 = x(2x-1),\ a_5 = (2x-1)^2,\ a_6 = (2x-1)(3x-2)$ , and in general, $a_{2n} = [(n-1)x - (n-2)][nx - (n-1)],$ $a_{2n+1} = [nx -(n-1)]^2$ . This we can easily show by induction: since $a_{2n} = 2a_{2n-1} - a_{2n-2} = 2[(n-1)x-(n-2)]^2 - [(n-2)x-(n-3)][(n-1)x-(n-2)] = []$ . The answer is $957 + 16$ .

2004 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2004 AIME II Problems/Problem 9

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