To show the sum of multiples of 3 between 1 and 100 is 1683.
The multiples of 3, in between 1 and 100 are 3, 6, 9, 12,… 99 which is an A.P.
In which a = 3, d = 6 – 3 = 3 and number of terms = 99/3 = 33.
Now sum of 33 terms of A.P.
3, 6, 9, …. 99 is
= 51 × 33
∴ Sum of multiples of ‘3’ = 1683
