Question

Find the sum of the first five multiples of $3$.

Answer 1

Find the sum of all the terms of the series.

The first five multiples of $3$are:$3,6,9,12,15$

Sum$=3+6+9+12+15=45$

Hence, the sum of the first five multiples of $3$ is $45$.

Alternatively,$3,6,9,12,15$

Here, $n=5,a=3$ and$l=15$

Sum of AP with first term$a$ and last term$l$is given by:

$$\begin{gathered} \Rightarrow S_n=\frac{n}{2}(a+l) \\ \Rightarrow S_n=\frac{5}{2}\times (3+15) \\ \Rightarrow S_n=\frac{5}{2}\times 18 \\ \Rightarrow S_n=5\times 9 \\ \Rightarrow S_n=45 \end{gathered}$$

Hence, the required sum is $45$.