Question

Find the sum of the first $8$ multiples of $3$

Answer 1

Step 1: Define the problem

The first eight multiples of $3$ can be expressed in the form of an arithmetic progression.
Arithmetic progression is a sequence where each term $a_{\mathrm{n}}$ is in the form of,

$a_{\mathrm{n}}=a+(n-1)d$

where, $a$ is the first term of the series, $d$ is the difference between any $n$^th and $(n-1)$^th term

Here, $a=3$, $d=3$ and $n=8$

Step 2: Apply the formula for the sum of AP

The sum ($S_n$) of an arithmetic progression is given as,

$S_n=\frac{n}{2}\left[2a+(n-1)d\right]$

Thus, the sum of first $8$ multiples of $3$ is,

$\begin{array}{rcl}{S}_n& =& \frac{8}{2}\left[2\times 3+(8-1)·3\right]\\ & =& 108\end{array}$

Thus, the sum of the first $8$ multiples of $3$ is $108$.