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

Answer

Number which are divisible by both 2 and 5 are the number which are divisible by 10.

Thus we need to find the number of natural numbers between 101 and 999 which are divisible by 10.

The first number between 101 and 999 which is divisible by 10 is 110.

And the last number between 101 and 999 which is divisible by 10 is 990.

Using the formula for Arithmetic progression where first term (a) = 110, and last term (Tn) = 990 and difference (d) =10.

             Tn = a+(n-1) x d

           990 = 110+(n-1) x 10 

           880 = (n-1) x 10

            88 = n-1

              n = 89

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

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Some More Questions From Arithmetic Progressions Chapter

Write first four terms of the AP, when the first term a and the common difference d are given as follows:
a = –2, d = 0

Write first four terms of the AP, when the first term a and the common difference d are given as follows:
a = 4, d = – 3

Write first four terms of the AP, when the first term a and the common difference d are given as follows:

a = – 1.25, d = – 0.25

For the following APs, write the first term and the common difference
3, 1, – 1, – 3, . . .

For the following APs, write the first term and the common difference
– 5, – 1, 3, 7, . . .

For the following APs, write the first term and the common difference
0.6, 1.7, 2.8, 3.9, . . .