, 2019

Solution:
Let $\{a_{n}\}$ have common difference $d$ and $\{b_{n}\}$ have common ratio $r$; for brevity, let $a_{1}=a$ and $b_{1}=b$. Then we have the equations $a b=20$, $(a+d) b r=19$, and $(a+2 d) b r^{2}=14$, and we want to maximize $(a+3 d) b r^{3}$.

The equation $(a+d) b r=19$ expands as $a b r+d b r=19$, or $20 r+b d r=19$ since $a b=20$. Similarly, $(20+2 b d) r^{2}=14$, or $10 r^{2}+b d r^{2}=7$. Multiplying the first equation by $r$ and subtracting the second, we get
$$10 r^{2}=19 r-7 \Longrightarrow (5 r-7)(2 r-1)=0$$
so either $r=\frac{7}{5}$ or $r=\frac{1}{2}$.

For each value of $r$, we have $b d=\frac{19-20 r}{r}=\frac{19}{r}-20$, so
$$(a+3 d) b r^{3}=(20+3 b d) r^{3}=\left(\frac{57}{r}-40\right) r^{3}=r^{2}(57-40 r)$$
The greater value of this expression is $\frac{37}{4}$, achieved when $r=\frac{1}{2}$.

Solution 2:
The key is to find a (linear) recurrence relation that the sequence $c_{n}=a_{n} b_{n}$ satisfies. Some knowledge of theory helps here: $c_{n}$ is of the form $s n r^{n}+t r^{n}$ for some constants $r, s, t$, so $\{c_{n}\}$ satisfies a linear recurrence relation with characteristic polynomial $(x-r)^{2}=x^{2}-2 r x+r^{2}$. That is,
$$c_{n}=2 r c_{n-1}-r^{2} c_{n-2}$$
for some constant $r$.

Taking $n=3$, we get $14=2 r \cdot 19-r^{2} \cdot 20$, which factors as $(5 r-7)(2 r-1)=0$, so either $r=\frac{7}{5}$ or $r=\frac{1}{2}$. Then
$$c_{4}=2 r c_{3}-r^{2} c_{2}=28 r-19 r^{2}$$
This expression is maximized at $r=\frac{14}{19}$, and strictly decreases on either side. Since $\frac{1}{2}$ is closer to $\frac{14}{19}$ than $\frac{7}{5}$, we should choose $r=\frac{1}{2}$, giving the answer $c_{4}=14-\frac{19}{4}=\frac{37}{4}$.