In a arithmetic sequence with a^1=12, d=3 which term is equal to 159?

Answer: n=50th term

Step-by-step explanation:

An = a1 + (n-1) d

An = nth term

A1 = first term

N = nth position

D = common difference

An = 159

A1 = 12

D = 3

N=?

Substitute the values

An=a1 + (n-1) d

159=12 + (n-1) 3

Collect like terms

159-12 = (n-1) 3

147=3n-3

147+3=3n

150=3n

N=150/3

N = 50

Therefore 50th term will give 156

50th term is 159

Step-by-step explanation:

nth term=159

a + (n - 1) * d = 159

12 + (n-1) * 3 = 159

(n - 1) * 3 = 159 - 12

(n - 1) * 3 = 147

n - 1 = 147/3

n - 1 = 49

n = 49 + 1

n = 50