For all n >=3

I will first prove that $a_{n} = a_{3}$ for $n \ge 3$ using induction.

Clearly $a_{3} = a_{3}$ , establishing the base case.

Now assume that $a_{k} = a_{3}$ for some $k \ge 3$ .

We have that $a_{k+1} = \dfrac{a_{1} + a_{2} + .... + a_{k-1} + a_{k}}{k}$ , $(A)$ .

But $a_{k} = \dfrac{a_{1} + a_{2} + ... + a_{k-1}}{k - 1}$ , and thus

$a_{1} + a_{2} + .... + a_{k-1} = (k - 1)a_{k}$ .

Substituting this result into equation $(A)$ we see that

$a_{k+1} = \dfrac{(k - 1)a_{k} + a_{k}}{k} = a_{k}$ ,

and so by the induction hypothesis $a_{k+1} = a_{3}$ , thus completing the proof by induction.

Thus $a_{3} = a_{9} = 99$ . But $a_{3} = \dfrac{a_{1} + a_{2}}{2}$ , so

$99 = \dfrac{19 + a_{2}}{2} \Longrightarrow a_{2} = 2*99 - 19 = \boxed{179}$ .

Let $S_n = a_1+a_2+a_3+...+a_n$ , then $a_n = \dfrac {S_{n-1}}{n-1}$ for $n \ge 3$ .

Now $a_3 = \dfrac {S_2}{2} = \dfrac {a_1+a_2}{2} \quad \Rightarrow a_1+a_2 = 2a_3 \quad \Rightarrow S_3 = a_1+a_2+a_3 = 3a_3$

$a_4 = \dfrac {S_3}{3} = \dfrac {3a_3}{3} = a_3\quad \Rightarrow S_4 = S_3 + a_4 = 3a_3 + a_3 = 4a_3$

Similarly, $a_5 = a_3$ and $S_5 = 5a_3... \quad \Rightarrow a_n = a_3$ for $n \ge 3$

$\quad \Rightarrow a_3 = a_9 = 99 \quad \Rightarrow a_3 = \dfrac {a_1+a_2}{2} = \dfrac {19+a_2}{2} = 99 \quad \Rightarrow a_2 = \boxed {179}$