Difference between revisions of "2012 AIME II Problems/Problem 2"

Revision as of 14:07, 4 July 2013

Problem 2

Two geometric sequences $a_1, a_2, a_3, \ldots$ and $b_1, b_2, b_3, \ldots$ have the same common ratio, with $a_1 = 27$ , $b_1=99$ , and $a_{15}=b_{11}$ . Find $a_9$ .

Solution

Call the common $r.$ Now since the $n$ th term of a geometric sequence with first term $x$ and common ratio $y$ is $xy^{n-1},$ we see that $a_1 \cdot r^{14} = b_1 \cdot r^{10} \longrightarrow r^4 = \frac{99}{27} = \frac{11}{3}.$ But $a_9$ equals $a_1 \cdot r^8 = a_1 \cdot (r^4)^2,$ so $a_9 = 27 \cdot \left(\frac{11}{3}\right)^2 = \boxed{363.}$

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 == See Also ==
 {{AIME box|year=2012|n=II|num-b=1|num-a=3}}
+{{MAA Notice}}
 [[Category:Introductory Algebra Problems]]

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Difference between revisions of "2012 AIME II Problems/Problem 2"

Revision as of 14:07, 4 July 2013

Problem 2

Solution

See Also