Question

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

Answer 1

Hint: We know that the number between $1$ and $500$ which are multiples of $2$ as well as of $5$ are $10$, $20$, $30$,..., $490$. This of the form arithmetic progression where first term $\left( a \right)$ is $10$, common difference $\left( d \right)$ is $10$ and last term is $490$.

Formula Used:
 ${a_n} = a + \left( {n - 1} \right)d$
Where, ${a_n}$ is the last term to find $n$ and $n$ is a number of elements.

Complete step by step solution
Given:
The numbers between $1$and $500$ which are multiples of $2$ as well as of $5$ that is which are multiples of $10$ are $10$,$20$, $30$,..., $490$.

The sequence is of the form of arithmetic progression where the first term is $10$ and the common difference is $10$ and the last term is $490$.

To find the number of elements in the sequence we use arithmetic progression we get,
${a_n} = 490$

We know the formula to find the ${n^{th}}$term in the arithmetic progression is,
${a_n} = a + \left( {n - 1} \right)d$

Where, $a$ is the first term and $n$ is the term and $d$ is the common difference.

On substituting the first term, last term and common difference we get,
$\begin{array}{c}
490 = 10 + \left( {n - 1} \right) \times 10\\
480 = \left( {n - 1} \right) \times 10\\
48 = n - 1\\
n = 49
\end{array}$

The number of terms in the sequence is $49$.

The main aim is to find the sum of integers for that we will use sum of $49$ arithmetic progression that is,
\[{S_n} = \dfrac{n}{2}\left[ {a + l} \right]\]

Where, $a$ is the first term and $l$ is the last term and $n$ is the number of terms in sequence.
On Substituting the values of first term and last term and $n$ we get,
$\begin{array}{c}
{S_n} = \dfrac{{49}}{2}\left[ {10 + 490} \right]\\
 = \dfrac{{49}}{2}\left[ {500} \right]\\
 = 12250
\end{array}$

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


Note: Since $2$and $5$ have no common factor we took multiples of $2 \times 5$ but if the two numbers have common factors take least common multiple for both. That are numbers common for the two numbers.