For this question, we use the formula for the nth term of an arithmetic sequence:
[tex]a_n=a_1+(n-1)d[/tex]Substituting a₁=12 and d=7 we get:
[tex]\begin{gathered} a_1=12+(1-1)7=12+0=12 \\ a_2=12+(2-1)7=12+7=19 \\ a_3=12+(3-1)7=12+14=26 \\ a_4=12+(4-1)7=12+21=33 \\ a_5=12+(5-1)7=12+28=40 \end{gathered}[/tex]