2019 AMC 10A Problem 15

The recursion has a product in the numerator and a difference in the denominator, which suggests taking reciprocals:
$$\frac{1}{a_n} = \frac{2a_{n-2} - a_{n-1}}{a_{n-2}\,a_{n-1}} = \frac{2}{a_{n-1}} - \frac{1}{a_{n-2}}.$$
Setting $b_n = 1/a_n$ gives $b_n = 2b_{n-1} - b_{n-2}$, i.e. $b_n - b_{n-1} = b_{n-1} - b_{n-2}$: the $b_n$ form an arithmetic sequence.

Its first two terms are $b_1 = 1$ and $b_2 = \tfrac73$, so the common difference is $\tfrac43$ and
$$b_n = 1 + \frac{4(n-1)}{3} = \frac{4n - 1}{3}.$$
Then $b_{2019} = \frac{4 \cdot 2019 - 1}{3} = \frac{8075}{3}$, so $a_{2019} = \frac{3}{8075}$. Since $8075 = 3 \cdot 2691 + 2$ is not a multiple of $3$, the fraction is reduced, and $p + q = 3 + 8075 = 8078$.

The answer is $\boxed{\textbf{(E)}\ 8078}$.

A recursion of the form $a_n = \frac{a_{n-1}a_{n-2}}{\alpha a_{n-2} + \beta a_{n-1}}$ becomes linear in the reciprocals, and a linear recursion with coefficients $2$ and $-1$ means constant differences. Whenever a recursion looks like a harmonic-mean formula, flip it. Computing $a_3 = \tfrac{3}{11}$ and $a_4 = \tfrac{3}{15}$ by hand shows the same pattern $a_n = \frac{3}{4n-1}$ and is a good sanity check.

Compute a few terms: $a_3 = \frac{1 \cdot \frac37}{2 - \frac37} = \frac{3}{11}$, $a_4 = \frac{\frac37 \cdot \frac{3}{11}}{\frac67 - \frac{3}{11}} = \frac{3}{15}$. Denominators $3, 7, 11, 15$ increase by $4$, so $a_n = \frac{3}{4n-1}$ and $a_{2019} = \frac{3}{8075}$.

• Mis-indexing the arithmetic sequence of reciprocals (using 4n+3 or 4n-5 instead of 4n-1), or grinding out fractions term by term instead of flipping the recursion.
• Writing the numerator as $1$ instead of $3$ (pattern $\frac{1}{\ldots}$), which leads to $p + q$ near $2020$ or $4039$.