Sequences and Series

The integers divisible by 5 from 1 to 100 are 5,10,15,20,25,...,100. Which are in A.P. with first term (a) = 5 and common difference (d)=5 ) last term (l)=100
So, total number of terms,
(n)‌=‌

l−a

d

+1=‌

100−5

5

+1=‌

95

5

+1
‌=19+1=20
∴Sn‌=‌

n

2

(a+l)
‌=10(5+100)=1050
Similarly, the integers divisible by 13 from 1 to 100 are 13,26,39,52,...,91 which are in AP with to a=13,d=13,l=91.
‌So,n=‌

l−a

d

+1=‌

91−13

13

+1=‌

78

13

+1
‌=6+1=7
‌∴‌‌Sn=‌

7

2

(13+91)=7×52=364
Now, the integer divisible by both 5 and 13 from 1 to 100 is 65 .
Hence, the sum of integer from 1 to 100 that are divisible by 5 or 13 is
=1050+364−65=1414−65=1349