2019/20 AHSMC Part 1 Problem 15

Since $a_{n+2} = 3(a_{n+1} - a_n)$ one obtains $a_2 = -3$, $a_3 = -9$, $a_4 = -18$, $\ldots$, and also, $$\begin{align*} a_{n+3} &= 3(a_{n+2} - a_{n+1}) = 3(2a_{n+1} - 3a_n), \\ a_{n+4} &= 3(a_{n+3} - a_{n+2}) = 3(3a_{n+1} - 6a_n), \\ a_{n+5} &= 3(a_{n+4} - a_{n+3}) = 3(3a_{n+1} - 9a_n), \\ a_{n+6} &= 3(a_{n+5} - a_{n+4}) = -3^3 a_n \end{align*}$$ hence $$a_{3+6n} = -3^3 a_{3+6(n-1)} = \cdots = (-1)^n 3^{3n} a_3 = (-1)^n 3^{3n}(-9) = (-1)^{n+1} 3^{3n+2}.$$ Thus, $$a_{2019} = a_{3 + 336 \cdot 6} = -3^{1010}.$$ The answer is (A).