2019 AMC 12A Problem 9

Question

A sequence of numbers is defined recursively by

a_1 = 1,

a_2 = \dfrac{3}{7},

and

a_n = \dfrac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}

for all

n \ge 3.

Then

a_{2019}

can be written as

\dfrac{p}{q},

where

p

and

q

are relatively prime positive integers. What is

p + q?

2020

4039

6057

6061

8078

Take reciprocals of both sides

Setting

b_n = \dfrac{1}{a_n}

gives

b_n = 2b_{n-1} - b_{n-2},

an arithmetic sequence

Taking reciprocals,

\begin{aligned} \dfrac{1}{a_n} &= \dfrac{2a_{n-2} - a_{n-1}}{a_{n-2}a_{n-1}} \\ &= \dfrac{2}{a_{n-1}} - \dfrac{1}{a_{n-2}}. \end{aligned}

Let

b_n = \dfrac{1}{a_n}.

Then

b_n = 2b_{n-1} - b_{n-2},

so

b_n

is arithmetic with

b_1 = 1,

b_2 = \dfrac{7}{3},

and common difference

\dfrac{4}{3}.

Thus

b_{2019} = 1 + 2018 \cdot \dfrac{4}{3} = \dfrac{8075}{3},

so

a_{2019} = \dfrac{3}{8075}.

Since these are relatively prime,

p + q = 8078.

Thus, the correct answer is E.

Problem 9 in other years