Find $N$ given a sequence $a_n$ 
In the sequence {$a_n$}, $$a_n=\frac{a_1+a_2+...+a_{n-1}}{n-1} \, , \quad n \ge 3$$
  If $a_1+a_2 \ne 0$ and the sum of the first $N$ terms is $12(a_1+a_2)$, find $N$.

Kind of lost on where to start with this one. My initial thought was, $$\sum^{N}_{n=3}\frac{a_1+a_2+...+a_{n-1}}{n-1}=12(a_1+a_2)$$ $$\sum^{N}_{n=3}\frac{a_1+a_2+...+a_{n-1}}{n-1}=\frac{1}{2}(a_1+a_2)+\frac{1}{3}(a_1+a_2+a_3)...$$but someting seems wrong here, or I do not know where to go from here.

\begin{align*}
  S_2 &= a_1+a_2 \\
  S_n &= a_1+a_2+\ldots+a_n \\
  a_{n+1} &= \frac{a_1+\ldots+a_{n}}{n} \\
  &= \frac{S_n}{n} \\
  S_{n+1} &= S_{n}+\frac{S_n}{n} \\
  &= \frac{n+1}{n} S_{n} \\
  &= \frac{n+1}{n} \cdot \frac{n}{n-1} \cdot \ldots \cdot \frac{4}{3} \cdot \frac{3}{2} S_{2} \\
  &= \frac{n+1}{2} S_2 \\
  S_{N} &= \frac{N}{2} S_2 \\
  &= \frac{N(a_1+a_2)}{2} \\
\end{align*}
Also $\forall n\ge 3$, $$a_{n}=\frac{a_1+a_2}{2}$$
