we have
a3=4
a4=7
a5=10
a6=13
a7=16
so
a4-a3=7-4=3
a5-a4=10-7=3
a6=a5=13-10=3
a7-a6=16-13=3
therefore
This is an arithmetic sequence, because the difference between consecutive terms is a constant called common difference d
d=3
the general equation is equal to
[tex]a_n=a_1+d(n-1)[/tex]
we have
d=3
so
For n=3
an=4
substitute and solve for a1
[tex]\begin{gathered} 4=a_1+3(3-1) \\ a_1=4-6=-2 \end{gathered}[/tex]
the general equation is
[tex]\begin{gathered} a_n=-2+3(n-1) \\ a_n=-2+3n-3 \\ a_n=3n-5 \end{gathered}[/tex]
the answer is
an=3n-5
