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
an = 994
n = ?
As we know,
an = 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.