Find the sum of integers from 1 to 100 that are divisible by 2 or 5. 

Solution:

The integers from 1 to 100 that are divisible by 2 are 2, 4, 6, …, 100.
This forms an A.P. with first term a = 2 and common difference d = 2.

100 = 2 + (n − 1)×2 ⇒ n = 50

∴ Sum = (n/2)[2a + (n − 1)d]

= (50/2)[2×2 + (50 − 1)×2] = 25[4 + 98] = 25×102 = 2550

The integers from 1 to 100 that are divisible by 5 are 5, 10, 15, …, 100.
This forms an A.P. with a = 5 and d = 5.

100 = 5 + (n − 1)×5 ⇒ 5n = 100 ⇒ n = 20

∴ Sum = (n/2)[2a + (n − 1)d]

= (20/2)[2×5 + (20 − 1)×5] = 10[10 + 95] = 10×105 = 1050

The integers divisible by both 2 and 5 are 10, 20, 30, …, 100.
This forms an A.P. with a = 10 and d = 10.

100 = 10 + (n − 1)×10 ⇒ 10n = 100 ⇒ n = 10

∴ Sum = (n/2)[2a + (n − 1)d]

= (10/2)[2×10 + (10 − 1)×10] = 5[20 + 90] = 5×110 = 550

Required sum = 2550 + 1050 − 550 = 3050

Hence, the sum of integers from 1 to 100 that are divisible by 2 or 5 is 3050.