Find the sum of integers from 1 to 100 that are divisible by 2 or 5.
The integers from 1 to 100 which are divisible by 2 are as follows: -
2, 4, 6… 100
Since the common difference between the consecutive terms is constant. Thus, the above sequence is an A.P. with both the first term and common difference equal to 2.
No. of terms of above A.P = 100/2 = 50
Last Term(l) = 100
Sum of n terms of an A.P(Sn) = (n/2)[a + l]
∴ S60 = (50/2)[2 + 100]
= 25 × 102
= 2550
The integers from 1 to 100 which are divisible by 5 are as follows: -
5, 10,… 100
Since the common difference between the consecutive terms is constant. Thus, the above sequence is an A.P. with both the first term and common difference equal to 5.
No. of terms of above A.P = 100/5 = 20
Last Term(l) = 100
Sum of n terms of an A.P(Sn) = (n/2)[a + l]
∴ S20 = (20/2)[5 + 100]
= 10 × 105
= 1050
The integers which are divisible by both 2 and 5 are as follows: -
10, 20, … 100
Since the common difference between the consecutive terms is constant. Thus, the above sequence is an A.P. with both the first term and common difference equal to 10.
No. of terms of above A.P = 100/10 = 10
Last Term(l) = 100
Sum of n terms of an A.P(Sn) = (n/2)[a + l]
∴ S10 = (10/2)[10 + 100]
= 5 × 110
= 550
∴Required sum = 2550 + 1050 – 550 = 3050
Thus, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.