Solution:

Write the common ratio of $a_1, a_2, a_3$ as $\frac{b}{a}$ in lowest terms, with $b \gt a \ge 1$ since the sequence increases. Because $a_3 = a_1 \left(\frac{b}{a}\right)^2$ is an integer and $\gcd(a, b) = 1,$ we get $a^2 \mid a_1;$ set $c = \frac{a_1}{a^2}.$ Then $a_1 = ca^2,$ $a_2 = cab,$ $a_3 = cb^2,$ and the arithmetic condition gives $a_4 = 2cb^2 - cab = cb(2b - a).$ Continuing, induction shows for every $k$ that $$a_{2k+1} = c\,\bigl(kb - (k-1)a\bigr)^2, \qquad a_{2k+2} = c\,\bigl(kb - (k-1)a\bigr) \bigl((k+1)b - ka\bigr).$$

In particular $a_{13} = c\,(6b - 5a)^2 = 2016 = 2^5 \cdot 3^2 \cdot 7.$ Let $N = 6b - 5a;$ then $N^2 \mid 2016,$ so $N \le 12.$ But $N = a + 6(b - a) \ge a + 6 \ge 7,$ and the only value in range with $N^2 \mid 2016$ is $N = 12,$ giving $c = \frac{2016}{144} = 14.$ From $6(b - a) = 12 - a$ we need $6 \mid a$ with $a \le 6,$ so $a = 6$ and $b = 7,$ which are coprime.

Therefore $a_1 = ca^2 = 14 \cdot 36 = 504.$ (Indeed the sequence begins $504, 588, 686, 784, 896, \ldots$ and reaches $a_{13} = 14 \cdot 12^2 = 2016.$)