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 sum of first five multiples of 3.
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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