[tex]\bf 7~~,~~\stackrel{7+6}{13}~~,~~\stackrel{13+6}{19}~~,~~\stackrel{19+6}{25}\qquad \impliedby \qquad \textit{common difference "d" is 6}[/tex]
we know all it's doing is adding 6 over again to each term to get the next one, so then
[tex]\bf \stackrel{\textit{Recursive Formula}}{\stackrel{\textit{nth term}}{f(n)}~~=~~\stackrel{\textit{the term before it}}{f(n-1)}~~~~\stackrel{\textit{plus 6}}{+~~~~6}}[/tex]
now for the explicit one
[tex]\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\[-0.5em] \hrulefill\\ a_1=7\\ d=6 \end{cases} \\\\\\ a_n=7+(n-1)6\implies a_n=7+6n-6\implies \stackrel{\textit{Explicit Formula}}{\stackrel{f(n)}{a_n}=6n+1} \\\\\\ therefore\qquad \qquad f(10)=6(10)+1\implies f(10)=61[/tex]
