Difference between revisions of "2012 AIME II Problems/Problem 2"
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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> | <!-- 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 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>. | ||
| + | |||
| + | == 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 \( a_1 = 27, b_1 = 99 \): | ||
| + | |||
| + | \( a_9 = \frac{99^2}{27} = 363 \). | ||
| + | |||
| + | <math>\boxed{363}</math> | ||
| + | |||
| + | ~Pinotation | ||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 19:59, 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
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. 
