If Na={an:n∈N} , then N3∩N4 =
N7
N12
N3
N4
From the given definition Na={an:n∈N} So we have, N3∩N4={3,6,9,12,15……}∩{4,8,12,16,20,……}
={12,24,36……}=N12
∴ N3∩N4 = N12