Q6 of 47 Page 1

Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.

Hence, numbers divisible by both 2 and 5 are 110, 120, 130, 140, 150, …., 990

⸫ The numbers are in arithmetic progression.


a (first term) = 110, d (common difference) = 10


Last term of an A.P. is given by,


l = a + (n – 1) d where, n = number of terms


⇒ 990 = 110 + (n – 1) 10


⇒ (n – 1) 10 = 990 – 110


⸫ n = 88 + 1 = 89


⸫ The number of terms = 89


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