Question

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

Answer 1

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.

Answer 2

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