**Solution:**

The three-digit numbers that are divisible by 7 are;

First number = 105

Second number = 105 + 7 = 112

Third number = 112 + 7 =119

Therefore, 105, 112, 119, …

This is an AP with the first term of 105 and a common difference of 7.

As we know, the largest possible three-digit number is 999.

When we divide 999 by 7, the remainder will be 5.

Therefore, 999 – 5 = 994 is the maximum possible three-digit number that is divisible by 7.

Now the series is as follows.

105, 112, 119, …, 994

Let 994 be the nth term of this AP.

First term, a = 105

Common difference, d = 7

a_{n} = 994

n = ?

As we know,

a_{n} = a + (n − 1)d

994 = 105 + (n − 1)7

889 = (n − 1)7

(n − 1) = 127

n = 128

Therefore, 128 three-digit numbers are divisible by 7.