Given :
All the natural numbers below 1000 that are multiples of 3 or 5 .
To Find :
The sum of all the multiples of 3 or 5 below 1000.
Solution :
Max multiple of 3 is 999 .
Max multiple of 5 id 995 .
So , number of multiple of 3 is :
[tex]999=a+(n-1)d\\\\999=3+3(n-1)\\\\n=333[/tex]
Similarly for 5 .
[tex]995=a+(n-1)d\\\\995=5+5(n-1)\\\\n=199[/tex]
Now , sum of all multiple of 3 is given by :
[tex]S_3=\dfrac{n}{2}(2a+(n-1)d)\\\\S_3=\dfrac{333\times (2\times 3+332\times 3)}{2}\\\\S_3=166833[/tex]
Also , sum of all multiple of 5 is :
[tex]S_5=\dfrac{n}{2}(2a+(n-1)d)\\\\S_5=\dfrac{199\times (2\times 5+198\times 5)}{2}\\\\S_5=99500[/tex]
Therefore , total sum :
[tex]T=S_3+S_5\\\\T=166833+99500\\\\T=266333[/tex]
Now , there are some common number which we add two times like :
15 , 30 , 60 ......
So , we should subtract the sum of all multiple of 15 from T .
Now , sum of all multiple of 15 is 33165 .
So ,
[tex]T=266333-33165\\\\T=233168[/tex]
Therefore , the sum of all the multiples of 3 or 5 below 1000 is 233168 .