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

By using the information in the question, find a equation in the y=mx+b form. Then, plug in 37 as the x value and solve for y.