Question

10 points for clear answer​

Answer 1

Answer:

89

Step-by-step explanation:

 As those numbers are divisible by both 2 and 5, it means they are divisible by 2*5 or 10.

   Using the properties of AP : nth term of AP is given by a + ( n - 1 )d, where a is the first term and d is the common difference.

Let the number of terms in this AP be n.

 Here,

a = first term = 110

l = last term = 990

co. diff. = 10

 

 990 = 110 + ( n - 1 )10

⇒ 990 - 110 = ( n - 1 )10

⇒ 880 = ( n - 1 )10

⇒ 88 = n - 1

⇒ 89 = n

  Hence the required number of terms is 89

Answer 2

there 88 numbers between 101 and 999 which are divisible by 2 and 5