Question

A sequence of numbers is defined recursively by $a_1 = 1,$ $a_2 = \frac{3}{7},$ and $a_n=\dfrac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}$ for all $n \geq 3.$ Then $a_{2019}$ can be written as $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. What is $p+q ?$

$2020$

$4039$

$6057$

$6061$

$8078$

Solution:

We can rewrite the recursive formula as $\begin{align*} \dfrac{1}{a_n} &= \dfrac{a_{n - 2} \cdot a_{n - 1}}{2a_{n - 2} - a_{n - 1}} \\&= \dfrac{2}{a_{n - 1}} - \dfrac{1}{a_{n - 2}}. \end{align*}$

This means that $\dfrac{1}{a_n} - \dfrac{1}{a_{n - 1}} = \dfrac{1}{a_{n - 1}} - \dfrac{1}{a_{n - 2}},$ which tells us that $\left\{\dfrac{1}{a_n}\right\}$ is an arithmetic sequence.

Using $a_1$ and $a_2,$ we get that the common difference is $\dfrac{1}{\frac{3}{7}} - \dfrac{1}{1} = \dfrac{7}{3} - 1 = \dfrac{4}{3}.$

From this, we get that $\begin{align*} \dfrac{1}{a_{2019}} &= 1 + 2018 \cdot \dfrac{4}{3} \\&= \dfrac{8075}{3}. \end{align*}$

$p + q$ is therefore $8075 + 3 = 8078.$

Thus, E is the correct answer.

2019 AMC 10A Problem 15

Solution:

Problem 15 in Other Years