First Term, a: 12; Common Difference, d=4; Last Term, a_{n}=248We need to find the number of terms (n) in the progression which ends at 248 .Using general term formula of Arithmetic Progression:​an​=a+(n−1)(d)248=12+(n−1)(4)248−12=4(n−1)236=4(n−1)59=n−1n=60​ Therefore, there are 60 terms, which means there are 60 multiples of 4 between 10 and 250 .