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

Revision as of 21:28, 14 March 2022

Problem 2

r.</math> 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 $a_1 \cdot r^{14} = b_1 \cdot r^{10} \implies r^4 = \frac{99}{27} = \frac{11}{3}.$ But $a_9$ equals $a_1 \cdot r^8 = a_1 \cdot (r^4)^2=27\cdot {\left(\frac{11}{3}\right)}^2=27\cdot \frac{121} 9=\boxed{363}$ .

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 == Problem 2 ==
-<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Two geometric sequences <math>a_1, a_2, a_3, \ldots</math> and <math>b_1, b_2, b_3, \ldots</math> have the same common ratio, with <math>a_1 = 27</math>, <math>b_1=99</math>, and <math>a_{15}=b_{11}</math>. Find <math>a_9</math>.<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
+<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude
 == Solution ==

2012 AIME II (Problems • Answer Key • Resources)
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Difference between revisions of "2012 AIME II Problems/Problem 2"

Revision as of 21:28, 14 March 2022

Problem 2

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