1) Considering that, let's find each term:
[tex]\begin{gathered} a_1=3 \\ a_n=a_{n-1}+4 \\ a_2=a_1+4\Rightarrow a_2=3+4=7 \\ a_3=a_2+4\Rightarrow a_3=7+4\text{ =11} \\ a_4=11+4\text{ }\Rightarrow a_4=15 \end{gathered}[/tex]2) So the sequence is
[tex](3,7,11,15,\ldots)[/tex]As each term, from the 2nd one is 4 units more that's why we can make it using a recursive formula
