2012 AIME II Problems/Problem 2
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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. 
