Question

The sum of all natural numbers between 1 and 100 which are multiples of 3 is.

• A. 1680
• B. 1683
• C. 1681
• D. 1682

Answer 1

Answer: (B)

1683

Answer 2

We get series 3, 6, 9, 12, ........ 99.

Here $n = \frac { 99 } { 3 } = 33 , a = 3 , d = 3$, therefore

$S = \frac { 33 } { 2 } \{ 2 \times 3 + ( 33 - 1 ) 3 \}$ $= \frac { 33 } { 2 } \times 102 = 33 \times 51 = 1683$ .