Difference between revisions of "2004 AIME II Problems/Problem 9"
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== Problem == | == Problem == | ||
| − | A sequence of positive integers with <math> a_1=1 </math> and <math> a_9+a_{10}=646 </math> 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 <math> n\ge1, </math> the terms <math> a_{2n-1}, a_{2n}, a_{2n+1} </math> are in geometric progression, and the terms <math> a_{2n}, a_{2n+1}, </math> and <math> a_{2n+2} </math> are in arithmetic progression. Let <math> a_n </math> be the greatest term in this sequence that is less than 1000. Find <math> n+a_n. </math> | + | A [[sequence]] of positive integers with <math> a_1=1 </math> and <math> a_9+a_{10}=646 </math> is formed so that the first three terms are in [[geometric sequence|geometric progression]], the second, third, and fourth terms are in [[arithmetic sequence|arithmetic progression]], and, in general, for all <math> n\ge1, </math> the terms <math> a_{2n-1}, a_{2n}, a_{2n+1} </math> are in geometric progression, and the terms <math> a_{2n}, a_{2n+1}, </math> and <math> a_{2n+2} </math> are in arithmetic progression. Let <math> a_n </math> be the greatest term in this sequence that is less than <math>1000</math>. Find <math> n+a_n. </math> |
== Solution == | == Solution == | ||
| − | {{ | + | Let <math>x = a_2</math>; then solving for the next several terms, we find that <math>a_3 = x^2,\ a_4 = x(2x-1),\ a_5 = (2x-1)^2,\ a_6 = (2x-1)(3x-2)</math>, and in general, <math>a_{2n} = [(n-1)x - (n-2)][nx - (n-1)],</math> <math> a_{2n+1} = [nx -(n-1)]^2</math>. This we can easily show by [[induction]]: since <math>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)] = []</math>. The answer is <math>957 + 16</math>. |
== See also == | == See also == | ||
{{AIME box|year=2004|n=II|num-b=8|num-a=10}} | {{AIME box|year=2004|n=II|num-b=8|num-a=10}} | ||
| + | |||
| + | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 18:13, 8 June 2008
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, ![$a_{2n} = [(n-1)x - (n-2)][nx - (n-1)],$](http://latex.artofproblemsolving.com/d/3/2/d3263dd5b7756a2079e7f0aca1d03dc7f3cb506b.png)
![$a_{2n+1} = [nx -(n-1)]^2$](http://latex.artofproblemsolving.com/5/c/e/5cebf83b5b9e1be2c8f07d899317b070d90e5550.png)
. This we can easily show by induction: since ![$a_{2n} = 2a_{2n-1} - a_{2n-2} = 2[(n-1)x-(n-2)]^2 - [(n-2)x-(n-3)][(n-1)x-(n-2)] = []$](http://latex.artofproblemsolving.com/e/6/0/e602ce9fc0c092b2a848dc4146760c9df83a7b16.png)
. The answer is 
.
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 | ||
