Answer :

We have to find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5 i.e. 10.

So, the AP is as follows: 110, 120, …, 990

From this, we get first term, a = 110, last term, l = 990 and common difference, d = 10

We know that l = a + (n – 1) d where n = number of terms

⇒ 990 = 110 + (n – 1) 10

⇒ 990 – 110 = 10n – 10

⇒ 880 + 10 = 10n

⇒ 890 = 10n

⇒ n = 890/ 10

⇒ n = 89

∴ There are 89 natural numbers divisible by 10 between 101 and 999.

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