# 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 integers. What is $p+q ?$

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

## Video Solution

Education, the Study of Everything

~ pi_is_3.14

## Solution 1 (Induction)

Using the recursive formula, we find $a_3=\frac{3}{11}$, $a_4=\frac{3}{15}$, 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 $a_m=\frac{3}{4m-1}$. Using the recursive formula, $$a_{m+1}=\frac{a_{m-1}\cdot a_m}{2a_{m-1}-a_m}=\frac{\frac{3}{4m-5}\cdot\frac{3}{4m-1}}{2\cdot\frac{3}{4m-5}-\frac{3}{4m-1}}=\frac{\left(\frac{3}{4m-5}\cdot\frac{3}{4m-1}\right)(4m-5)(4m-1)}{\left(2\cdot\frac{3}{4m-5}-\frac{3}{4m-1}\right)(4m-5)(4m-1)}=\frac{9}{6(4m-1)-3(4m-5)}=\frac{3}{4(m+1)-1},$$ so our induction is complete.

## Solution 2

We have $\frac{1}{a_n} = \frac{2a_{n-2}-a_{n-1}}{a_{n-2} \cdot a_{n-1}}=\frac{2}{a_{n-1}}-\frac{1}{a_{n-2}}$, in other words, $\frac{1}{a_n}-\frac{1}{a_{n-1}} = \frac{1}{a_{n-1}}-\frac{1}{a_{n-2}}$. So $\{\frac{1}{a_n}\}$ is an arithmetic sequence with step size $\frac{7}{3}-1=\frac{4}{3}$, which means $\frac{1}{a_n} = 1+2018 \cdot \frac{4}{3} = \frac{8075}{3}$. Since the numerator and the denominator are relatively prime, the answer is $\boxed{\textbf{(E) } 8078}$.

-eric2020 (modified by Dolphindesigner)

## Solution 3

It seems reasonable to transform the equation into something else. Let $a_{n}=x$, $a_{n-1}=y$, and $a_{n-2}=z$. Therefore, we have $$x=\frac{zy}{2z-y}$$ $$2xz-xy=zy$$ $$2xz=y(x+z)$$ $$y=\frac{2xz}{x+z}$$ Thus, $y$ is the harmonic mean of $x$ and $z$. This implies $a_{n}$ is a harmonic sequence or equivalently $b_{n}=\frac{1}{a_{n}}$ is arithmetic. Now, we have $b_{1}=1$, $b_{2}=\frac{7}{3}$, $b_{3}=\frac{11}{3}$, and so on. Since the common difference is $\frac{4}{3}$, we can express $b_{n}$ explicitly as $b_{n}=\frac{4}{3}(n-1)+1$. This gives $b_{2019}=\frac{4}{3}(2019-1)+1=\frac{8075}{3}$ which implies $a_{2019}=\frac{3}{8075}=\frac{p}{q}$. $p+q=\boxed{\textbf{(E) } 8078}$ ~jakeg314

## Solution 4 (Arithmetic Sequence)

Notice that$$a_n = \frac{1}{\frac{2}{a_{n-1}} - \frac{1}{a_{n-2}}}.$$Therefore,$$\frac{1}{a_n} = \frac{2}{a_{n-1}} - \frac{1}{a_{n-2}}, \ \ \implies \ \ \frac{\frac{1}{a_n} + \frac{1}{a_{n-2}}}{2} = \frac{1}{a_{n-1}}.$$Therefore, the sequence $b_n = \frac{1}{a_n}$ is an arithmetic sequence. Notice that the common diffrence of $b$ is $\frac{4}{3},$ and therefore$$b_{2019} = b_1 + 2018 \bigg(\frac{5}{3}\bigg) = 1 + 2018 \bigg(\frac{4}{3} \bigg) = \frac{8075}{3}.$$Therefore, we see that $a_{2019} = \frac{3}{8075},$ so that $p + q = \boxed{\text{(E) } 8078}.$

~Professor-Mom

Note: This is similar to solutions #2 and #3, although you can notice that in #2's case the new sequence $B$ actually forms an arithmetic sequence.