Difference between revisions of "2010 AMC 12B Problems/Problem 20"
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== Solution == | == Solution == | ||
− | By defintion, we have <math>\cos^2x=\sin x \tan x</math>. Since <math>\tan x=\frac{\sin x}{\cos x}</math>, we can rewrite this as <math>\cos^3x=\sin^2x</math>. | + | By defintion of a geometric sequence, we have <math>\cos^2x=\sin x \tan x</math>. Since <math>\tan x=\frac{\sin x}{\cos x}</math>, we can rewrite this as <math>\cos^3x=\sin^2x</math>. |
The common ratio of the sequence is <math>\frac{\cos x}{\sin x}</math>, so we can write | The common ratio of the sequence is <math>\frac{\cos x}{\sin x}</math>, so we can write |
Revision as of 19:21, 18 August 2011
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
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 |