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
