Difference between revisions of "2010 AMC 12B Problems/Problem 20"
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<cmath>a_7=\frac{\cos^3x}{\sin^3x}=\frac{1}{\sin x}</cmath> | <cmath>a_7=\frac{\cos^3x}{\sin^3x}=\frac{1}{\sin x}</cmath> | ||
<cmath>a_8=\frac{\cos x}{\sin^2 x}=\frac{1}{\cos^2 x}</cmath> | <cmath>a_8=\frac{\cos x}{\sin^2 x}=\frac{1}{\cos^2 x}</cmath> | ||
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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>. |
Revision as of 02:43, 7 February 2012
Problem
A geometric sequence has , , and for some real number . For what value of does ?
Solution
By defintion of a geometric sequence, we have . Since , we can rewrite this as .
The common ratio of the sequence is , so we can write
Since , we have , which is making our answer .
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 |