The sequence $a_{1},a_{2},a_{3},\ldots$ satisfies $a_{1} = 19,a_{9} = 99$, and, for all $n\geq 3$, $a_{n}$ is the arithmetic mean of the first $n - 1$ terms.  Find $a_2$.
$\textrm{(A)} \ 29 \qquad \textrm{(B)} \ 59 \qquad \textrm{(C)} \ 79 \qquad \textrm{(D)} \ 99 \qquad \textrm{(E)} \ 179$
Let $m$ be the arithmetic mean of $a_1$ and $a_2$. We can then write $a_1=m-x$ and $a_2=m+x$ for some $x$.
By definition, $a_3=m$.
Next, $a_4$ is the mean of $m-x$, $m+x$ and $m$, which is again $m$. 
Realizing this, one can easily prove by induction that $\forall n\geq 3;~ a_n=m$.
It follows that $m=a_9=99$. From $19=a_1=m-x$ we get that $x=80$. And thus $a_2 = m+x = \boxed{(E)  179}$.