The sum of the first five multiples of 3 is | vedclass

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(D) The first five multiples of $3$ are $3, 6, 9, 12,$ and $15$.This forms an Arithmetic Progression $(AP)$ where the first term $a = 3$,the common di

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$45$

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(D) The first five multiples of $3$ are $3, 6, 9, 12,$ and $15$.This forms an Arithmetic Progression $(AP)$ where the first term $a = 3$,the common difference $d = 3$,and the number of terms $n = 5$.The sum of the first $n$ terms of an $AP$ is given by the formula $S_n = \frac{n}{2} [2a + (n - 1)d]$.Substituting the values:$S_5 = \frac{5}{2} [2(3) + (5 - 1)3]$$S_5 = \frac{5}{2} [6 + 4(3)]$$S_5 = \frac{5}{2} [6 + 12]$$S_5 = \frac{5}{2} [18]$$S_5 = 5 	imes 9 = 45$.

The sum of the first five multiples of $3$ is

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