Solution:

Numbers that are divisible by 2 and 5 are likewise divisible by 10.
Now, the numbers divisible by 10 integers between 101 and 999 are 110, 120, 130,…, 990.
Clearly, these figures are in AP. Then,

Here, a =110 and d=120-110=10 Let this AP contains n terms. Then, a_n = 990

110+(n-1)x10=990                     [a. = a +(n-1)d]

10n +100 = 990

lOn = 990 —100 = 890

n= 89

Hence, there are 89 numbers between 101 and 999 which are divisible by both 2 and 5.