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(B) The natural numbers between $1$ and $100$ which are multiples of $3$ form an arithmetic progression: $3, 6, 9, \dots, 99$.Here,the first term $a =

Vedclass · Accepted Answer

$1683$

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(B) The natural numbers between $1$ and $100$ which are multiples of $3$ form an arithmetic progression: $3, 6, 9, \dots, 99$.Here,the first term $a = 3$,the common difference $d = 3$,and the last term $l = 99$.To find the number of terms $n$,we use the formula $l = a + (n - 1)d$:$99 = 3 + (n - 1)3$$96 = (n - 1)3$$32 = n - 1$$n = 33$.The sum $S_n$ of an arithmetic progression is given by $S_n = \frac{n}{2}(a + l)$:$S_{33} = \frac{33}{2}(3 + 99)$$S_{33} = \frac{33}{2}(102)$$S_{33} = 33 	imes 51 = 1683$.

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

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