First 3-digit number divisible by 6 is 102 and last 3-digit number divisible by 6 is 996.
Difference between two consecutive numbers divisible by 6 is 6.
So 3-digit numbers divisible by 6 are 102,108,114, ......., 996.
This is an Arithmetic Progression in which a = 102, d = 6 and l = 996.
where a = First Number , l = Last Number and d = difference of two consecutive numbers.
Let the number of terms be n. So Last term = tn
Then tn = 996
Use the formula for n terms of arithmetic progression.
∴ a + ( n - 1) x d = 996
⇒ 102 + (n - 1) x 6 = 996
⇒ 6(n - 1) = 894
⇒ (n - 1) = 149
⇒ n = 150
∴ Numbers of terms = 150
