Difference between revisions of "2012 AIME II Problems/Problem 2"
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== Solution 1 == | == Solution 1 == | ||
Call the common ratio <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>. | Call the common ratio <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>. | ||
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== Solution 2== | == Solution 2== | ||
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~Pinotation | ~Pinotation | ||
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| + | ==Video Solution== | ||
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| + | https://youtu.be/V2X9hz6DuUw | ||
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| + | ~Lucas | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 19:57, 18 August 2025
Problem 2
Two geometric sequences 
and 
have the same common ratio, with 
, 
, and 
. Find 
.
Solution 1
Call the common ratio 
Now since the 
th term of a geometric sequence with first term 
and common ratio 
is 
we see that 
But equals 
.
Solution 2
Let the ratio be \( r \). From \( \frac{a_{15}}{b_{11}} = \frac{a_{15}}{b_{11}} \):
\( a_1 r^{14} = b_1 r^{10} \implies a_1 r^4 = b_1 \).
Notice how \( a_5 = a_1 r^4 = b_1 \).
Then
\( a_9 = a_5 r^4 = b_1 r^4 = \frac{b_1^2}{a_1} \).
Plug in \( a_1 = 27, b_1 = 99 \):
\( a_9 = \frac{99^2}{27} = 363 \).
~Pinotation
Video Solution
~Lucas
Video Solution
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
