2010 AMC 12B Problem 20

Problem 20 of 25HarderAlgebra Geometry

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?

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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.

More practice

Concepts: geometric sequence · trigonometric identity

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