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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 $a_n=1+\cos x$?

The AMC website has a solution to this, and I wish I could post that here. I would like to know how else this could be solved, not using their solution process.

The solution they used was to show $\cos^3 x=\sin^2 x$ and then multiply by the common ratio $\cos x/\sin x$ until $$a_8= \frac 1 {\cos^2 x}$$ is reached. They then, after rewriting $\sin^2 x$ with cosine squared, proceeded to show that $$\frac 1{\cos^2 x}=1+\cos x$$

Is there any other method? Someone mentioned finding a geometric series within a geometric series, which I could not produce.

1 Answers1

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As the sequence is geometric, we may write it as $$ \sin x, q\,\sin x, q^2\,\sin x, \ldots $$ Then $q=\cos x/\sin x$. We also know that $$ q^2=\frac{\tan x}{\sin x}=\frac1{\cos x}, \ q^3=q\,q^2=\frac{\cos x}{\sin x}\,\frac1{\cos x}=\frac1{\sin x}. $$ So we want $a_n=1+\cos x$, i.e. $q^{n-1}\sin x=1+\cos x$. Then $$ q^{n-1}=\frac1{\sin x}+\frac{\cos x}{\sin x}=q^3+q. $$ Cancelling one $q$ we get $$ q^{n-2}=q^2+1=\frac{\cos^2x}{\sin ^2x}+1=\frac1{\sin^2x}=\frac1{\sin x}\,\frac1{\sin x}=q^3\,q^3=q^6. $$ So $n=8$.

Martin Argerami
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  • Wonderful! I only wish I could think like that. What led you to try that method? –  Nov 09 '13 at 23:32
  • If you think about it, I just wrote what the geometric progression was, and I started writing the powers of $q$. Then I wrote the equation $a_n=1+\cos x$ using the information I had about the first 3 powers. – Martin Argerami Nov 09 '13 at 23:50