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 :
![999=a+(n-1)d\\\\999=3+3(n-1)\\\\n=333](https://img.qammunity.org/2021/formulas/mathematics/high-school/fhuad7bnwzlupml30pt3p1t6p66bz0cq1n.png)
Similarly for 5 .
![995=a+(n-1)d\\\\995=5+5(n-1)\\\\n=199](https://img.qammunity.org/2021/formulas/mathematics/high-school/dkvgghemgkbwhk0rcos69wgkm9x9u4tps9.png)
Now , sum of all multiple of 3 is given by :
![S_3=(n)/(2)(2a+(n-1)d)\\\\S_3=(333* (2* 3+332* 3))/(2)\\\\S_3=166833](https://img.qammunity.org/2021/formulas/mathematics/high-school/vsqv4er4og5yy0ld64mrmmoizbebjf7ybt.png)
Also , sum of all multiple of 5 is :
![S_5=(n)/(2)(2a+(n-1)d)\\\\S_5=(199* (2* 5+198* 5))/(2)\\\\S_5=99500](https://img.qammunity.org/2021/formulas/mathematics/high-school/e5kunr013aa7928j70unxly2adugddpq8j.png)
Therefore , total sum :
![T=S_3+S_5\\\\T=166833+99500\\\\T=266333](https://img.qammunity.org/2021/formulas/mathematics/high-school/wnnjrbww48sr5nqv40bguwxkb63ga0i7ns.png)
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 ,
![T=266333-33165\\\\T=233168](https://img.qammunity.org/2021/formulas/mathematics/high-school/mz25cn0vopr6nvxb4uxernhnu62k85qkqd.png)
Therefore , the sum of all the multiples of 3 or 5 below 1000 is 233168 .