Difference between revisions of "2012 AIME II Problems/Problem 2"
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== 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> | ||
== 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}} | ||
Revision as of 16:38, 31 March 2012
Problem 2
Two geometric sequences 
and 
have the same common ratio, with 
, 
, and 
. Find 
.
Solution
Call the common 
Now since the 
th term of a geometric sequence with first term 
and common ratio 
is 
we see that 
But equals 
so 
See also
| 2012 AIME II (Problems • Answer Key • Resources) | ||
| 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 | ||
