The initial three-digit number divisible by \(7\) is \(105\). The subsequent numbers are \(105 + 7 = 112\), \(112 + 7 = 119\), and so on. This forms an arithmetic progression (A.P.) with the first term \(a = 105\) and a common difference \(d = 7\). The largest three-digit number is \(999\). Dividing \(999\) by \(7\) yields a remainder of \(5\). Therefore, \(999 - 5 = 994\) is the largest three-digit number divisible by \(7\). The sequence of three-digit numbers divisible by \(7\) is \(105, 112, 119, …, 994\). To find the number of terms (\(n\)) in this A.P., we set the last term \(a_n = 994\). Using the formula for the nth term of an A.P., \(a_n = a + (n − 1) d\), we have \(994 = 105 + (n − 1)7\). Subtracting \(105\) from both sides gives \(889 = (n − 1)7\). Dividing by \(7\), we get \(n − 1 = 127\). Thus, \(n = 128\). There are \(128\) three-digit numbers divisible by \(7\).