The sum to $n$ terms of the $A.P$, whose  $k^{th}$ term  $5k+1$ is

$a_k=5k+1$

$a+(k-1)d=5k+1$

$a+kd-d=5k+1$

Compare co – efficient of  $k\Rightarrow d=5$

$\begin{array}{l}a-d=1\\ a-5=1\\ a=6\\ S_n=\frac{n}{2}[2a+(n-1)d]\\ =\frac{n}{2}[2\left(6\right)+(n-1)5]\\ =\frac{n}{2}[12+5n-5]\\ S_n=\frac{n}{2}(5n+7)\end{array}$

(or)

$\begin{array}{l}{a}_n=S_{n+1}\\ S_n=\Sigma a_n\\ =\Sigma (5n+1)\\ =\frac{5\left(n\right)(n+1)}{2}+n\\ =\frac{5n^2+5n+2n}{2}\\ =\frac{5n^2+7n}{2}\\ =\frac{n}{2}(5n+7)\end{array}$