Answer :
I'm guessing you mean
[tex]S_{k+1} = S_k + a_{k+1}[/tex]
and that [tex]S_k[/tex] is the sum
[tex]S_k = 6 + 12 + 18 + \ldots + 6k = 6 (1 + 2 + 3 + \ldots + k) = 3 k (k+1)[/tex]
using the well-known formula
[tex]\displaystyle \sum_{i=1}^n i = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2[/tex]
Then by substitution,
[tex]S_{k+1} = 6 (1 + 2 + 3 + \cdots + k + (k+1)) = 3 (k+1) (k+2)[/tex]
and hence
[tex]a_{k+1} = S_{k+1} - S_k[/tex]
[tex]a_{k+1} = 3 (k+1) (k+2) - 3k (k+1)[/tex]
[tex]a_{k+1} = 3 (k+1) \bigg((k+2) - k\bigg)[/tex]
[tex]\implies \boxed{a_{k+1} = 6 (k + 1)}[/tex]