Sum of all natural numbers between 1 and 100 which are multiples of 3 is

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8. Sequences and Series

easy

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

A

$1680$

B

$1683$

C

$1681$

D

$1682$

Solution

(b) We get series $3, 6, 9, 12, …….. 99.$

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

$S = \frac{{33}}{2}\left\{ {2 \times 3 + (33 – 1)3} \right\}$

$ = \frac{{33}}{2} \times 102 = 33 \times 51 = 1683$.

Standard 11

Mathematics

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