[tex]a_{n}=a_{n-1}+(n-1)\times4[/tex]
1) To get to know the answer to this problem, we need to consider the first term 15 the common difference obtained by the difference between the terms:
[tex]\begin{gathered} 19-15=4 \\ 23-19=4 \\ 27-23=4 \end{gathered}[/tex]
2) Therefore, we can write the recursive formula resorting to the previous term, i.e. our recursive formula is valid for the second term and so forth and so on.
[tex]\begin{gathered} a_n=a_{n-1}+(n-1)d \\ \\ a_n=15+(2-1)4 \\ \\ a_2=15+4\Rightarrow a_2=19 \end{gathered}[/tex]
3) Thus, the answer is:
[tex]a_n=a_{n-1}+(n-1)\cdot4[/tex]
