Solution:

Let $a$ be the first term and $r$ be the common ratio. Thus, $\frac{a}{1-r} = 2020$, or $a = 2020(1 - r)$.

We also have $a r^{2} - a = a r^{3} - a r^{2}$. Since the sum of the geometric series is nonzero, $a \neq 0$, and so we have $r^{2} - 1 = r^{3} - r^{2}$, or $r^{3} - 2 r^{2} + 1 = 0$.

Since the sum of the geometric series is finite, $r$ cannot be $1$, so $r^{3} - 2 r^{2} + 1 = (r - 1)(r^{2} - r - 1) = 0$ implies $r^{2} - r - 1 = 0$.

Solving this equation gives $r = \frac{1 - \sqrt{5}}{2}$ (since $|r| < 1$). This gives us $a = 2020 \left( \frac{1 + \sqrt{5}}{2} \right ) = 1010(1 + \sqrt{5})$.