Difference between revisions of "2010 AMC 12B Problems/Problem 20"
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==Solution 2== | ==Solution 2== | ||
− | Notice that the common ratio is <math>r=\frac{\cos(x)}{\sin(x)}</math>; multiplying it to <math>\tan(x)=\frac{\sin(x)}{\cos(x)}</math> gives <math>a_4=1</math>. Then, working backwards we have <math>a_3=\frac{1}{r}</math>, <math>a_2=\frac{1}{r^2}</math> and <math>a_1=\frac{1}{r^3}</math>. Now notice that since <math>a_1=\sin(x)</math> and <math> | + | Notice that the common ratio is <math>r=\frac{\cos(x)}{\sin(x)}</math>; multiplying it to <math>\tan(x)=\frac{\sin(x)}{\cos(x)}</math> gives <math>a_4=1</math>. Then, working backwards we have <math>a_3=\frac{1}{r}</math>, <math>a_2=\frac{1}{r^2}</math> and <math>a_1=\frac{1}{r^3}</math>. Now notice that since <math>a_1=\sin(x)</math> and <math>a_2=\cos(x)</math>, we need <math>a_1^2+a_2^2=1</math>, so <math>\frac{1}{r^6}+\frac{1}{r^4}=\frac{r^2+1}{r^6}=1\implies r^2+1=r^6</math>. Dividing both sides by <math>r^2</math> gives <math>1+\frac{1}{r^2}=r^4</math>, which the left side is equal to <math>1+\cos(x)</math>; we see as well that the right hand side is equal to <math>a_8</math> given <math>a_4=1</math>, so the answer is <math>\boxed{E}</math>. - mathleticguyyy |
== 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 15:32, 19 August 2019
Contents
Problem
A geometric sequence has , , and for some real number . For what value of does ?
Solution
By the 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 .
Solution 2
Notice that the common ratio is ; multiplying it to gives . Then, working backwards we have , and . Now notice that since and , we need , so . Dividing both sides by gives , which the left side is equal to ; we see as well that the right hand side is equal to given , so the answer is . - mathleticguyyy
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.