Two geometric sequences a_1, a_2, a_3, … and b_1, b_2, b_3, … have the same common ratio, with a_1 = 27, b_1 = 99, and a_15 = b_11. Find a_9.

2012 AIME II Problem 2

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Two geometric sequences

a_1,

a_2,

a_3,

\ldots

and

b_1,

b_2,

b_3,

\ldots

have the same common ratio, with

a_1 = 27,

b_1 = 99,

and

a_{15} = b_{11}.

Find

a_9.

Answer

With common ratio

r,

the condition

a_{15} = b_{11}

says

27r^{14} = 99r^{10}

You only need

r^4,

because

a_9 = 27r^8

Let

r

be the shared common ratio. Then

a_{15} = 27r^{14}

and

b_{11} = 99r^{10},

27r^{14} = 99r^{10}

gives

r^4 = \frac{99}{27} = \frac{11}{3}.

Therefore

\begin{aligned} a_9 &= 27r^8 = 27\left(\frac{11}{3}\right)^2 \\ &= 27 \cdot \frac{121}{9} = 3 \cdot 121 = 363. \end{aligned}