Difference between revisions of "2010 AMC 12B Problems/Problem 20"
m (→Solution) |
Cellsecret (talk | contribs) (→Problem) |
||
(4 intermediate revisions by 2 users not shown) | |||
Line 1: | Line 1: | ||
== Problem== | == Problem== | ||
A geometric sequence <math>(a_n)</math> has <math>a_1=\sin x</math>, <math>a_2=\cos x</math>, and <math>a_3= \tan x</math> for some real number <math>x</math>. For what value of <math>n</math> does <math>a_n=1+\cos x</math>? | A geometric sequence <math>(a_n)</math> has <math>a_1=\sin x</math>, <math>a_2=\cos x</math>, and <math>a_3= \tan x</math> for some real number <math>x</math>. For what value of <math>n</math> does <math>a_n=1+\cos x</math>? | ||
− | |||
<math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8</math> | <math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8</math> | ||
Line 20: | Line 19: | ||
− | 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>. |
+ | |||
+ | ==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>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}} |
Latest revision as of 19:52, 24 June 2022
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.