Answer:
1683
first multiple (a1) of 3 between 1 and 100 is 3, last multiple (an) is 99, using that, find n first (no of terms or n) with d = 3
an=a1+(n-1)d
99=3+(n-1)3
96=3(n-1)
n-1=32
n=33
to find the sum, use the formula
Sn=n[(a1+an)/2]
Sn=33[(3+99)/2]
Sn=33(102/2)
Sn=33(51)
Sn=1683
