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# Choose the correct answer from the given four options:

The sum of first five multiples of 3 is

**(A)** 45

**(B)** 55

**(C)** 65

**(D)** 75

Given:

First five multiples of 3.

To do:

We have to find the

Solution:

First five multiples of 3 forms an AP, where

$a_1 = a= 3, a_2=6$

Common difference $d =a_2-a_1$

$=6-3$

$=3$

We know that,

Sum of the $n$ terms of an AP is $S_n=\frac{n}{2}[2a+(n-1)d]$

Therefore,

$S_{5}= \frac{5}{2}[2(3)+(5-1)(3)]$

$= \frac{5}{2}(6+4(3))$

$= \frac{5}{2}(18)$

$=5\times 9$

$=45$

The sum of the first 5 multiples of 3 is $45$.

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