Since, the number is divisible by both 2 and 5, means it must be divisible by 10.
In the given numbers, first number that is divisible by 10 is 110.
Next number is 110 + 10 = 120.
The last number that is divisible by 10 is 990.
Thus, the progression will be 110, 120, ..., 990.
All the terms are divisible by 10, and thus forms an A.P. having first term as 110 and the common difference as 10.
We know that, nth term = an = a + (n − 1)d
According to the question,
990 = 110 + (n − 1)10
⇒ 990 = 110 + 10n − 10
⇒ 10n = 990 − 100
⇒ 10n = 890
⇒ n = 89
Thus, the number of natural numbers between 101 and 999 which are divisible by both 2 and 5 is 89.