Find the second, fourth, and tenth terms of the sequence described the the rule.

a(n)=4+(n-1)(8)

It's easy, just plug the values requested in n & solve:

A(n)=4+(n-1)(8)

second, n=2 ==> A₂ = 4 +(2-1)(8) = 12
fourth, n=4 ===> A₄ = 4 +(4-1)(8) = 20
tenth, n=10 ==>  A₁₀ = 4 + (10-1)(8) = 72

This an Arithmetic progression:
with 1st term a₁ =4 & d ( common difference) =8

Ibelieve that c. recording would work best.