Difference between revisions of "2012 AIME II Problems/Problem 2"
m (→Solution) |
(→Video Solution) |
||
| (10 intermediate revisions by 5 users not shown) | |||
| Line 3: | Line 3: | ||
== 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} \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>. |
| + | |||
| + | ==Video Solution== | ||
| + | |||
| + | https://youtu.be/V2X9hz6DuUw | ||
| + | |||
| + | ~Lucas | ||
| + | |||
| + | ==Video Solution== | ||
| + | |||
| + | https://youtu.be/Zfx5rP4GP6w | ||
== See Also == | == See Also == | ||
Latest revision as of 14:51, 21 August 2024
Problem 2
Two geometric sequences 
and 
have the same common ratio, with 
, 
, and 
. Find 
.
Solution
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 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
