2010 AMC 12B Problems/Problem 20

Problem

A geometric sequence $(a_n)$ has $a_1=\sin x$ , $a_2=\cos x$ , and $a_3= \tan x$ for some real number $x$ . For what value of $n$ does $a_n=1+\cos x$ ?

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$

Solution

By the defintion of a geometric sequence, we have $\cos^2x=\sin x \tan x$ . Since $\tan x=\frac{\sin x}{\cos x}$ , we can rewrite this as $\cos^3x=\sin^2x$ .

The common ratio of the sequence is $\frac{\cos x}{\sin x}$ , so we can write

$a_1= \sin x$ $a_2= \cos x$ $a_3= \frac{\cos^2x}{\sin x}$ $a_4=\frac{\cos^3x}{\sin^2x}=1$ $a_5=\frac{\cos x}{\sin x}$ $a_6=\frac{\cos^2x}{\sin^2x}$ $a_7=\frac{\cos^3x}{\sin^3x}=\frac{1}{\sin x}$ $a_8=\frac{\cos x}{\sin^2 x}=\frac{1}{\cos^2 x}$

Since $\cos^3x=\sin^2x=1-\cos^2x$ , we have $\cos^3x+\cos^2x=1 \implies \cos^2x(\cos x+1)=1 \implies \cos x+1=\frac{1}{\cos^2 x}$ , which is $a_8$ , making our answer $8 \Rightarrow \boxed{E}$ .

Solution 2

Notice that the common ratio is $r=\frac{\cos(x)}{\sin(x)}$ ; multiplying it to $\tan(x)=\frac{\sin(x)}{\cos(x)}$ gives $a_4=1$ . Then, working backwards we have $a_3=\frac{1}{r}$ , $a_2=\frac{1}{r^2}$ and $a_1=\frac{1}{r^3}$ . Now notice that since $a_1=\sin(x)$ and $a_2=\cos(x)$ , we need $a_1^2+a_2^2=1$ , so $\frac{1}{r^6}+\frac{1}{r^4}=\frac{r^2+1}{r^6}=1\implies r^2+1=r^6$ . Dividing both sides by $r^2$ gives $1+\frac{1}{r^2}=r^4$ , which the left side is equal to $1+\cos(x)$ ; we see as well that the right hand side is equal to $a_8$ given $a_4=1$ , so the answer is $\boxed{E}$ . - mathleticguyyy

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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All AMC 12 Problems and Solutions

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2010 AMC 12B Problems/Problem 20

Contents

Problem

Solution

Solution 2

See also