Solution:
We all know that,
Multiple of 2 + Multiple of 5 – LCM Multiple (2, 5) = Multiples of 2 or 5
Multiple of 2 + Multiple of 5 – Multiple of LCM (10) = Multiples of 2 or 5
List of multiple of 2 from 1 to 500 + List of multiple of 5 from 1 to 500 – List of multiple of 10 from 1 to 500 = Multiples of 2 or 5 from 1 to 500
Required sum 
Now consider the first series,
2, 4, 6, …., 500
The first term, a = 2
The common difference, d = 2
Let no of terms be n
Sum of an AP, ![{{S}_{n~}}=\text{ }\left( n/2 \right)\text{ }[\text{ }a\text{ }+\text{ }{{a}_{n}}]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8e5e624399550d65293ccfe825775ef5_l3.png "Rendered by QuickLaTeX.com")
Let S1 be ,the sum of this AP
![{{S}_{1~}}=\text{ }{{S}_{250}}~=\text{ }\left( 250/2 \right)\text{ }\times \left[ 2+500 \right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-031d24aa2cab802d8cb38af33eb4bafd_l3.png "Rendered by QuickLaTeX.com")
… (1)
Now consider the second series,
5, 10, 15, …., 500
The first term, a = 5
The common difference, d = 5
Let no of terms be n
Using the nth term formula
Sum of an AP,
Let S2, be the sum of this AP
![{{S}_{2~}}=\text{ }{{S}_{100}}~=\text{ }\left( 100/2 \right)\text{ }\times \left[ 5+500 \right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7cac474d448ac0a92141bfc4addd8cb2_l3.png "Rendered by QuickLaTeX.com")
… (2)
Now consider the third series,
10, 20, 30, …., 500
The first term, a = 10
The common difference, d = 10
Let no of terms be n
Sum of an AP,
Let S3, be the sum of this AP
![{{S}_{3~}}=\text{ }{{S}_{50}}~=\text{ }\left( 50/2 \right)\text{ }\times \text{ }\left[ 2+510 \right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2cb436a02348e330e5e3d7a289f8bb73_l3.png "Rendered by QuickLaTeX.com")
… (3)
As a result, the required Sum, 
S 
