jmc

For $n\ge 3,$ we have that $$a_n=\frac{a_1+a_2+\cdots +a_{n-1}}{n-1},$$or $$(n-1)a_n=a_1+a_2+\cdots +a_{n-1}.$$Likewise, $$n{a}_{n+1}=a_1+a_2+\cdots +a_{n-1}+a_n.$$Subtracting these equations, we get $$n{a}_{n+1}-(n-1)a_n=a_n,$$so $n{a}_{n+1}=n{a}_n.$ Then $a_{n+1}=a_n.$

This means that the terms $a_3,$ $a_4,$ $a_5,$ $\dots$ are all equal. In particular, $a_3=99,$ so $$\frac{19+a_2}{2}=99.$$We find $a_2=\boxed{179}.$