The multiples of 4 between 12 and 250 form the AP: 16, 20, 24, …, 248 with $a = 16$, $d = 4$, $a_n = 248$.
$248 = 16 + (n-1) \times 4 \Rightarrow 232 = (n-1) \times 4 \Rightarrow n = 59$
Answer: (A) 59
Source: Exercise 5.2, Q.14, Chapter 5
The key is identifying the first multiple of 4 strictly between 12 and 250 (which is 16, not 12) and the last (which is 248, not 252). Applying $a_n = a + (n-1)d$ gives $n = 59$. Students often err by including 12 or 252, getting 60 or 61. The textbook Exercise 5.2 Q.14 asks a similar question ("between 10 and 250"), giving 59 — confirming option A here.