11. An infinite geometric series has sum 2020. If the first term, the third term, and the fourth term form an arithmetic sequence, find the first term.
Answer: $1010(1+\sqrt{5})$
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 $r^{3}-2 r^{2}+1=(r-1)\left(r^{2}-r-1\right)=0$ implies $r^{2}-r-1=0$. Solving this equation gives $r=\frac{1-\sqrt{5}}{2}$ (since $|r|<1$ ). This gives us $a=2020\left(\frac{1+\sqrt{5}}{2}\right)=1010(1+\sqrt{5})$.