Question

find the sum of all natural numbers is that less than hundred and divisible by 3

Answer 1

your answer is 5049 it is correct

Answer 2

a=t1=3,t2=6,t3=9......tn=99
d=t2-t1=6-3=3
tn=a+(n-1)d
99=3+(n-1)3
99=3+3n-3
99=3n
n=99/3
n=33
Sn=2a+(n-1)d
S33=2(3)+33-1(3)
S33=6+32(3)
=6+96
S33=102
Hope it is helpful