Average of first nine integral multiples of 3 is

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4.Average

easy

The average of the first nine integral multiples of $3$ is

A

$12$

B

$15$

C

$18$

D

$21$

Solution

Sum of the $9$ integral multiple of $3 .$

$=\frac{9}{2}[2 \times 3+(9-1) 3] \quad\left[\begin{array}{l}\text { Using Formula } \\ S_{n}=\frac{n}{2}[2 a+(n-1) d]\end{array}\right]=135$

Average $=\frac{135}{9}=15$

Standard 13

Quantitative Aptitude

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