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

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== Problem 2 ==
 
== Problem 2 ==
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>.
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<!-- 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>
  
 
== Solution ==
 
== Solution ==
Call the common <math>r.</math> Now since the <math>n</math>th term of a geometric sequence with first term <math>x</math> and common ratio <math>y</math> is <math>xy^{n-1},</math> we see that <math>a_1 \cdot r^{14} = b_1 \cdot r^{10} \longrightarrow r^4 = \frac{99}{27} = \frac{11}{3}.</math> But <math>a_9</math> equals <math>a_1 \cdot r^8 = a_1 \cdot (r^4)^2,</math> so <math>a_9 = 27 \cdot \left(\frac{11}{3}\right)^2 = \boxed{363.}</math>
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Call the common <math>r.</math> Now since the <math>n</math>th term of a geometric sequence with first term <math>x</math> and common ratio <math>y</math> is <math>xy^{n-1},</math> we see that <math>a_1 \cdot r^{14} = b_1 \cdot r^{10} \implies r^4 = \frac{99}{27} = \frac{11}{3}.</math> But <math>a_9</math> equals <math>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}</math>.
  
 
== See Also ==
 
== See Also ==
 
{{AIME box|year=2012|n=II|num-b=1|num-a=3}}
 
{{AIME box|year=2012|n=II|num-b=1|num-a=3}}
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{{MAA Notice}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 14:02, 28 June 2021

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

See Also

2012 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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