# 2019 AMC 10A Problems/Problem 15

The following problem is from both the 2019 AMC 10A #15 and 2019 AMC 12A #9, so both problems redirect to this page.

## Problem

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

$\textbf{(A) } 2020 \qquad\textbf{(B) } 4039 \qquad\textbf{(C) } 6057 \qquad\textbf{(D) } 6061 \qquad\textbf{(E) } 8078$

## Solution

Using the recursive formula, we find $a_3=\frac{3}{10}$, $a_4=\frac{3}{14}$, and so on. It appears that $a_n=\frac{3}{4n-1}$, for all $n$. Setting $n=2019$, we find $a_{2019}=\frac{3}{8075}$, so the answer is $\boxed{\textbf{(E) }8078}$.

To prove this formula, we use induction. We are given that $a_1=1$ and $a_2=\frac{3}{7}$, which satisfy our formula. Now assume the formula holds true for all $n\le m$ for some positive integer $m$. By our assumption, $a_{m-1}=\frac{3}{4m-5}$ and $a_m=\frac{3}{4m-2}$. Using the recursive formula, $$a_{m+1}=\frac{a_{m-1}\cdot a_m}{2a_{m-1}-a_m}=\frac{\frac{3}{4m-5}\cdot\frac{3}{4m-2}}{2\cdot\frac{3}{4m-5}-\frac{3}{4m-2}}=\frac{(\frac{3}{4m-5}\cdot\frac{3}{4m-2})(4m-5)(4m-2)}{(2\cdot\frac{3}{4m-5}-\frac{3}{4m-2})(4m-5)(4m-2)}=\frac{9}{6(4m-2)-3(4m-5)}=\frac{3}{4m+1},$$ so our induction is complete.