Solution 1. 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 $r=\frac{1}{2}$.