Question

What is the 32nd term of an arithmetic sequence with a first term of 7 and a common difference of 4?

Select one:
a. 119
b. 123
c. 127
d. 131

Answer 1

Answer:

a₃₂ = 131

Step-by-step explanation:

the nth term of an arithmetic sequence is

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

here a₁ = 7 and d = 4 , then

a₃₂ = 7 + (31 × 4) = 7 + 124 = 131