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, $a_{2n} = f(n-1)f(n)$ , $a_{2n+1} = f(n)^2$ , where $f(n) = nx - (n-1)$ .^[1]

From $a_9 + a_{10} = f(4)^2 + f(4)f(5) = (4x-3)(9x-7) = 646 = 2\cdot 17 \cdot 19$ , we find that by either the quadratic formula or trial-and-error/modular arithmetic that $x=5$ . Thus $f(n) = 4n+1$ , and we need to find the largest $n$ such that either $f(n)^2\, \mathrm{or}\, f(n)f(n-1) < 1000$ . This happens with $f(7)f(8) = 29 \cdot 33 = 957$ , and this is the $2(8) = 16$ th term of the sequence.

The answer is $957 + 16 = \boxed{973}$ .

^ We can show this by simultaneous induction: since $\begin{align*}a_{2n} &= 2a_{2n-1} - a_{2n-2} = 2a_{2(n-1)+1} - a_{2(n-1)} \\ &= 2f(n-1)^2 - f(n-2)f(n-1) \\ &= f(n-1)[2f(n-1) - f(n-2)] \\ &= f(n-1)[(2n-2-n+2)x-(2n-4-n+3)] \\ &= f(n-1)f(n) \end{align*}$ and $\begin{align*}a_{2n+1} &= \frac{a_{2n}^2}{a_{2n-1}} = \frac{f(n-1)^2f(n)^2}{f(n-1)^2} = f(n)^2 \\ \end{align*}$

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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