Question

find the sum of all the integers from 100 and 200 which are divisible by 3

Answer 1

102 105 108..........................198

Its an AP

a = 102   d = 3   l = 198      An = a + (n-1)d

l = 102 + (n-1)3           n = 33

Sn = n/2 (a + l)

33/2 (102 + 198)

Sn = 4950

Answer 2

Hey dear here is your answer
_________________________

102,105,108........................198
let this the AP
a=102,d=3,an=198
an=a+(n-1)d
198=102+(n-1)3
198-102=(n-1)3
96=(n-1)3
96/4=(n-1)
24=(n-1)
24+1=n
n=25
sn=n/2(a+l)
sn= 25/2(102+198)
=25/2×300
=25× 150
=3750
so, the sum of all integers between 100 and 200 that are divisible by 3 is 3750

hope it may help you!!!!!!!!