Question

Find the average of first $20$ multiples of $3$.

Answer 1

Finding the average of first $20$ multiples of $3$:

First $20$ multiples of $3$ are:

$\left(3\times 1\right),\left(3\times 2\right),\left(3\times 3\right),\left(3\times 4\right),\left(3\times 5\right),\left(3\times 6\right),.....,\left(3\times 20\right)$

It can also be written as:

$\frac{3\left(1+2+3+4+5+6+7+8+9+....+20\right)}{20}$

Also, the sum of first $n$ natural numbers is $\frac{n\left(n+1\right)}{2}$.

So,

$\begin{array}{rcl}\frac{3\left(1+2+3+4+5+6+7+8+9+....+20\right)}{20}& =& \frac{3\left(20\right)\left(20+1\right)}{2\left(20\right)}[\because n=20]\\ & =& \frac{63}{2}\\ & =& 31.5\end{array}$

Hence, the average of first $20$ multiples of $3$ is $31.5$.