Question

Find the sum of those integers between $1$and$500$ which are multiples of $2$as well as of $5$.

Answer 1

Solve for the series of common multiples of $2$ and $5$ and calculate the sum

Step 1: Find out the series formed by the multiples

$LCMof(2,5)=10$

$\therefore$Multiples of $2$ as well as$5$ between $1$ and $500$ are $10,20,30,\dots ,490.$

The series $10,20,30\dots ,490$ is an A.P. with a common difference, $d=10$

First-term, $a=10$ and last term $490$

Step 2: Find the total number of terms

We know that the $n^{th}$ term of an A.P. is given by the formula

$a_n=a+(n–1)d$ where,

$a=$ first term

$a_n$ is the $nth$ term

$d$ is the common difference

$$\begin{gathered} \Rightarrow a_n=a+(n–1)\times d \\ \Rightarrow 490=10+(n–1)\times 10 \\ \Rightarrow 480=(n–1)10 \\ \Rightarrow n–1=48 \\ \Rightarrow n=49 \end{gathered}$$

Step 3: Find the sum of arithmetic progression.

The sum of an AP is given by

$$\begin{gathered} S_n=\frac{n}{2}[a+a_n], \\ {S}_n=\frac{49}{2}\times [10+490] \\ {S}_n=\frac{49}{2}\times \left[500\right] \\ {S}_n=49\times 250 \\ {S}_n=12250 \end{gathered}$$

Hence, the sum of those integers between $1$and $500$which are multiples of$2$ as well as of $5$is$12250$.