Difference between revisions of "2004 AIME II Problems/Problem 9"
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{{note|1}} We can show this by simultaneous [[induction]]: since | {{note|1}} We can show this by simultaneous [[induction]]: since | ||
− | < | + | <cmath>\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) \\ | &= 2f(n-1)^2 - f(n-2)f(n-1) \\ | ||
&= f(n-1)[2f(n-1) - f(n-2)] \\ | &= f(n-1)[2f(n-1) - f(n-2)] \\ | ||
&= f(n-1)[(2n-2-n+2)x-(2n-4-n+3)] \\ | &= f(n-1)[(2n-2-n+2)x-(2n-4-n+3)] \\ | ||
− | &= f(n-1)f(n) \end{align*}</ | + | &= f(n-1)f(n) \end{align*}</cmath> |
and | and | ||
− | < | + | <cmath>\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*}</ | + | \end{align*}</cmath> |
== See also == | == See also == |
Latest revision as of 19:55, 13 March 2015
Problem
A sequence of positive integers with and 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 the terms are in geometric progression, and the terms and are in arithmetic progression. Let be the greatest term in this sequence that is less than . Find
Solution
Let ; then solving for the next several terms, we find that , and in general, , , where .^{[1]}
From , we find that by either the quadratic formula or trial-and-error/modular arithmetic that . Thus , and we need to find the largest such that either . This happens with , and this is the th term of the sequence.
The answer is .
^ We can show this by simultaneous induction: since and
See also
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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.