Question

Find the sum of first $40$ positive integers divisible by $6$.

Answer 1

Find the sum of terms:

We have to find the sum of the first $40$ positive integers divisible by$6$

The positive integers that are divisible by $6$ are $6,12,18,24\dots .$

We know, that this series forms an A.P. whose first term is$6$and the common difference is$6.$

The first term,$a=6$

The common difference, difference is $6$

$S_{40}=?$

By the formula of the sum of $n$ terms, we know,

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

Therefore, putting $n=40$, we get,

$$\begin{gathered} \Rightarrow S_{40}=\frac{40}{2}[2\times 6+(40-1)\times 6] \\ \Rightarrow S_{40}=20\times [12+39\times 6] \\ \Rightarrow S_{40}=20\times (12+234) \\ \Rightarrow S_{40}=20\times 246 \\ \Rightarrow S_{40}=4920 \end{gathered}$$

Hence, the sum of the first $40$ positive integers divisible by $6$ is $4920$.