If the sum of the series$2, 5, 8, 11, ...$is $60100$, then $n$, the number of terms, is

We have,

$\begin{array}{l}\frac{n}{2}[2\left(2\right)+(n-1\left)3\right]=60100\\ \\ n(3n+1)=120200\\ 3n^2+n-120200=0\\ 3n^2-600n+601n+120200=0\\ (n-200)(3n+601)=0\Rightarrow n=200\end{array}$