Question

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

Answer 1

The list of numbers from $1$ to $500$ that are the multiples of $2$ as well as $5$ are :$10,20,30,40,.......,490$

Now, the above list forms an AP in which first term, $a=10$ and common difference, $d=10$

Let $a_n=490$

$\Rightarrow a+(n-1)d=490$

$\Rightarrow 10+(n-1)10=490$

$\Rightarrow (n-1)10=480$

$\Rightarrow n-1=48$

$\Rightarrow n=49$

Now, Sum of first ${}^{\prime }{n}^{\prime }$ terms of AP is

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

$S_{49}=\frac{49}{2}[2\times 10+48\times 10]$

$=\frac{49}{2}(20+480)$

$=49\times 250$

$=12250$