2019 AMC 10A Problem 15

Taking reciprocals in the recursive formula gives $$\begin{aligned} \frac1{a_n} &=\frac{2a_{n-2}-a_{n-1}}{a_{n-2}a_{n-1}}\\ &=\frac2{a_{n-1}}-\frac1{a_{n-2}}. \end{aligned}$$ 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 $$\dfrac{1}{\frac{3}{7}} - \dfrac{1}{1} = \dfrac{7}{3} - 1 = \dfrac{4}{3}.$$ Therefore $$\frac1{a_{2019}}=1+2018\cdot\frac43=\frac{8075}{3},$$ so $a_{2019}=\frac3{8075}.$ $p + q$ is therefore $$8075 + 3 = 8078.$$ Thus, E is the correct answer.