Find the sum of all integers from 1 and 100 that are divisible by 2 or 5

Solution:

The integers from 1 and 100 that are divisible by 2 are 2, 4, 6, ....., 100

This forms an A.P with both the first term and common difference equal to 2.

Therefore,

a_n = a + (n - 1) d

100 = 2 + (n - 1) 2

100 = 2 + 2n - 2

⇒ 2n = 100

⇒ n = 50

Therefore,

2, 4, 6, ....., 100 = 50/2 [2 (2) + (50 - 1)(2)]

= 50/2 [4 + 98]

= 25 x 102

= 2550

The integers from 1 to 100 that are divisible by 5 are 5, 10, 15, ...., 100

This forms an A.P with both the first term and common difference equal to 5

Therefore,

a_n = a + (n - 1) d

100 = 5 + (n - 1)5

100 = 5 + 5n - 5

⇒ 5n = 100

⇒ n = 20

Therefore,

5, 10, 15, ...., 100 = 20/2 [2 (5) + (20 - 1)(5)]

= 10 [10 + (19)5]

= 10 [10 + 95]

= 10 x 105

= 1050

The integers, which are divisible by both 2 and 5 are 10, 20, 30, ...., 100

This forms an A.P with both the first term and common difference equal to 10.

Therefore,

100 = 10 + (n - 1)10

⇒ 10n = 100

⇒ n = 10

Hence,

10, 20,30, ...., 100 = 10/2 [2 (10) + (10 - 1)(10)]

= 5 [20 + 90]

= 5 x 110

= 550

Therefore, required sum = 2550 + 1050 - 550 = 3050

Thus, the sum of all integers from 1 and 100 that are divisible by 2 or 5 is 3050

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NCERT Solutions Class 11 Maths Chapter 9 Exercise ME Question 5

Find the sum of all integers from 1 and 100 that are divisible by 2 or 5

Summary:

The solution to the sum of all integers from 1 and 100 that are divisible by 2 or 5 is 3050