2010 AMC 12B Problem 20

Below is the professionally curated solution for Problem 20 of the 2010 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 12B solutions, or check the answer key.

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Concepts:geometric sequence trigonometric identity

Difficulty rating: 2240

20.

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 $a_n=1+\cos x?$

$4$

$5$

$6$

$7$

$8$

Solution:

The common ratio is $r=\dfrac{a_2}{a_1}=\cot x.$ Then $a_4=a_3\cdot r=\tan x\cot x=1.$

From $a_3=a_1r^2,$ we get $\tan x=\sin x\cot^2 x=\dfrac{\cos^2 x}{\sin x},$ so $\sin^2 x=\cos^3 x,$ i.e. $(\cos^2 x)(1+\cos x)=1.$

Hence $1+\cos x=\dfrac{1}{\cos^2 x}.$ Also $r^2=\dfrac{\cos^2 x}{\sin^2 x}=\dfrac{\cos^2 x}{\cos^3 x}=\dfrac{1}{\cos x},$ so $r^4=\dfrac{1}{\cos^2 x}=1+\cos x.$

Therefore $1+\cos x=a_4\cdot r^4=a_8,$ so $n=8.$

Thus, the correct answer is E.

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