2010 AMC 12B Problems/Problem 20: Difference between revisions
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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 | |
