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 $p + q?$

$2020$

$4039$

$6057$

$6061$

$8078$

Solution:

Taking reciprocals, $\dfrac{1}{a_n} = \dfrac{2a_{n-2} - a_{n-1}}{a_{n-2}a_{n-1}} = \dfrac{2}{a_{n-1}} - \dfrac{1}{a_{n-2}}.$

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.

2019 AMC 12A Problem 9

Solution:

Problem 9 in Other Years