The first multiple of \(4\) greater than \(10\) is \(12\), followed by \(16\). This forms an arithmetic progression (A.P.): \(12, 16, 20, 24, …\). All terms are divisible by \(4\), with the first term \(a = 12\) and common difference \(d = 4\).
Dividing \(250\) by \(4\) yields a remainder of \(2\). Therefore, \(250 − 2 = 248\) is divisible by \(4\).
The series up to \(248\) is: \(12, 16, 20, 24, …, 248\).
Let \(248\) be the nth term (\(a_n\)). Using the A.P. formula \(a_n = a + (n-1)d\):
\(248 = 12 + (n-1)4\)
\(248 - 12 = (n-1)4\)
\(236 = (n-1)4\)
\(\frac {236}{4} = n-1\)
\(59 = n-1\)
\(n = 59 + 1 = 60\).
Thus, there are \(60\) multiples of \(4\) between \(10\) and \(250\).