If the sum of the series 2, 5, 8, 11, ……….  n terms is 60100, then find the value  of n.

Here first term is a=2 and common difference d=3.

Hence sum of n terms of A.P. is (n/2)(4+3(n-1))=60100

$\begin{array}{l}\Rightarrow 3n^2+n-120200=0\\ \Rightarrow (n-200)(3n+601)=0\Rightarrow n=200\end{array}$