If the second term of an arithmetic sequence is 5 and the fourth term is 12, what is the 37th term?

Aritmetic sequence
an=a1+d(n-1)
a1=first term
d=common differnce
an=nth term

given
a2=5
a4=12

so
a2=a1+d(2-1)=5
a4=a1+d(4-1)=12

a1+d=5
a1+3d=12
so 12-5=(a1+d)-(a1+3d)=2d=7
2d=7
divide by 2
d=3.5

a2=a1+3.5(1)
5=a1+3.5
minus 3.5 both sides
1.5=a1

an=1.5+3.5(n-1)
37th term
a37=1.5+3.5(37-1)
a37=1.5+3.5(36)
a37=1.5+126
a37=127.5

37th term is 127.5

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.

Dm me pls mike and