Difference between revisions of "2010 AMC 12B Problems/Problem 20"
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Since <math>\cos^3x=\sin^2x=1-\cos^2x</math>, we have <math>\cos^3x+\cos^2x=1 \implies \cos^2x(\cos x+1)=1 \implies \cos x+1=\frac{1}{\cos^2 x}</math>, which is <math>a_8</math> making our answer <math>8 \Rightarrow \boxed{E}</math>. | Since <math>\cos^3x=\sin^2x=1-\cos^2x</math>, we have <math>\cos^3x+\cos^2x=1 \implies \cos^2x(\cos x+1)=1 \implies \cos x+1=\frac{1}{\cos^2 x}</math>, which is <math>a_8</math> making our answer <math>8 \Rightarrow \boxed{E}</math>. | ||
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== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=19|num-a=21|ab=B}} | {{AMC12 box|year=2010|num-b=19|num-a=21|ab=B}} |
Revision as of 21:27, 31 May 2011
Problem
A geometric sequence has , , and for some real number . For what value of does ?
Solution
By defintion, we have . Since , we can rewrite this as .
The common ratio of the sequence is , so we can write
We can conclude that the sequence from to repeats.
Since , we have , which is making our answer .
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See also
2010 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |