jmc

$\text{Solution 1: Find the first term and common ratio.}$

Let the first term be $a$ and the common ratio be $r.$ Because the second term is $-2,$ we have $ar=-2.$ Because the fifth term is $16,$ we have $a{r}^4=16.$ Dividing this by $ar=-2,$ we have $r^3=-8.$ Therefore, $r=-2.$ So, $a=-2/r=1.$ Therefore, the fourteenth term is $a{r}^{13}=(1)(-2{)}^{13}=\boxed{-8192}.$

$\text{Solution 2: Use our understanding of geometric sequences.}$

To get from the second term to the fifth term, we multiply by the common ratio, $r,$ three times. Therefore, multiplying $-2$ by $r^3$ gives us $16.$ So, $r^3=-8.$ Rather than finding $r,$ we note that to get to the fourteenth term from the fifth term, we multiply by $r$ nine times, which is the same as multiplying by $r^3$ three times. So, the fourteenth term is $16(-8{)}^3=\boxed{-8192}.$