Assuming t(1) is our first term, we can say:
t(2)=t(1)+d
t(3)=t(2)+d
t(4)=t(3)+d
This can go on forever, we can then establish the recursive formula which states:
t(n+1)=t(n)+d
or
t(n)=t(n-1)+d
Explicit formula:
t(2)=t(1)+d
t(3)=t(2)+d, but t(2)=t(1)+d
t(3)=t(1)+d+d
t(3)=t(1)+2d
t(4)=t(1)+3d
t(n)=t(1)+(n-1)d
We can calculate d for these equations:
t(8)=t(1)+7d
67=4+7d
7d=63
d=9
Explicit relation:
t(n)=4+9(n-1)
t(n)=9n-5
Recursive relation:
t(n)=t(n-1)+9
t(500)=9(500)-5=4500-5=4495