We have to find the number of three-digit numbers which are divisible by 7.

aₙ = a + (n - 1)d is the nth term of AP.

Here, aₙ is the nth term, a is the first term, d is the common difference and n is the number of terms.

First three-digit number that is divisible by 7 = 105

Next number = 105 + 7 = 112

Therefore, the series becomes 105, 112, 119, ...

Thus, 105, 112, 119, ... is forming an A.P. having the first term as 105 and a common difference of 7.

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

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

Hence the final sequence is as follows:

105, 112, 119, ..., 994

Let 994 be the nth term of this A.P.

a = 105

d = 7

aₙ = 994

n = ?

We know that the nth term of an A.P. is, aₙ = a + (n - 1)d

994 = 105 + (n - 1)7

889 = (n - 1)7

n - 1 = 889/7

n - 1 = 127

n = 127 + 1

n = 128

There are 128 three-digit numbers that are divisible by 7