The sum of all natural numbers between 1 and | vedclass

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Question

(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 	imes 3 +

Vedclass · Accepted Answer

$1683$

Vedclass · Answer

(b) We get series $3, 6, 9, 12, ........ 99.$

Here $n = \frac{{99}}{3} = 33,\;a = 3,\;d = 3$, therefore

$S = \frac{{33}}{2}\left\{ {2 	imes 3 + (33 - 1)3} ight\}$

$ = \frac{{33}}{2} 	imes 102 = 33 	imes 51 = 1683$.

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

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