AIME 2012 II · 第 2 题

Solution 1

Call the common ratio $r.$ Now since the $n$th term of a geometric sequence with first term $x$ and common ratio $y$ is $xy^{n-1},$ we see that $a_1 \cdot r^{14} = b_1 \cdot r^{10} \implies r^4 = \frac{99}{27} = \frac{11}{3}.$ But $a_9$ equals $a_1 \cdot r^8 = a_1 \cdot (r^4)^2=27\cdot {\left(\frac{11}{3}\right)}^2=27\cdot \frac{121} 9=\boxed{363}$.

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 $a_1 = 27, b_1 = 99$:

$a_9 = \frac{99^2}{27} = 363$.

$\boxed{363}$

~Pinotation

Video Solution

https://youtu.be/V2X9hz6DuUw

~Lucas