Question

If sum of $n$ terms of a series is given by $S_n=3n^2+3n$, then $6^{th}$ term of the series is

Answer 1

Given : $S_n=3n^2+3n$
Now,
$T_n=S_n-S_{n-1}\Rightarrow T_n=3n^2+3n-\left(3\right(n-1{)}^2+3(n-1))\Rightarrow T_n=3n^2+3n-(3(n^2-2n+1)+3n-3)\Rightarrow T_n=3n^2+3n-(3n^2-3n)\Rightarrow T_n=6n\therefore T_6=36$