Question

The sum of all natural number between 1 to 100 which are multiple of 3

Answer 1

3+6+9+12+...............+99 = 1683. It's a question of A.P . It is done by using fomula.
Sn = n/2 (2a + nd-d )
n = number of terms
a = first term
d = common difference.
a = 3
d = 3
n = 33
Sn = 33/2× {2×3+[33-1]×3}
33/2×[6+32×3]
33/2×[6+96]
33/2×102
33×51
1683.

Answer 2

S=3,6,9,.....96,99
A=3,d=3
An=a+(n-1)d
99=3+(n-1)3
99=3+3n-3
99=3n
n=33.
By using formula..
Sn=n/2(a+l) {a=first term.l=last term}
=33/2(3+99)
=33/2(102)
=33×51
Sn =1683