2012 AIME II Problem 2

Below is the professionally curated solution for Problem 2 of the 2012 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AIME II solutions, or check the answer key.

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Concepts:geometric sequencealgebraic manipulation

Difficulty rating: 1750

2.

Two geometric sequences a1,a2,a3,a_1, a_2, a_3, \ldots and b1,b2,b3,b_1, b_2, b_3, \ldots have the same common ratio, with a1=27,a_1 = 27, b1=99,b_1 = 99, and a15=b11.a_{15} = b_{11}. Find a9.a_9.

Solution:

Let rr be the shared common ratio. Then a15=27r14a_{15} = 27r^{14} and b11=99r10,b_{11} = 99r^{10}, so 27r14=99r1027r^{14} = 99r^{10} gives r4=9927=113.r^4 = \frac{99}{27} = \frac{11}{3}.

Therefore a9=27r8=27(113)2=271219=3121=363.a_9 = 27r^8 = 27\left(\frac{11}{3}\right)^2 = 27 \cdot \frac{121}{9} = 3 \cdot 121 = 363.

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