Find the sum of all three digits multiples of 4. Class 10 CBSE Arthmetic progressions.

Multiples of 4 can be written as 4k, for some integer k.The first three digits multiple of 4 is 4*25=100, and the last is 4*249=996.So, the sum of all the three digits multiples of 4 is[tex]\displaystyle \sum_{k=25}^{249}4k = 4\sum_{k=25}^{249}k[/tex]We know how to sum the first [tex]N[/tex] integers, but we need to sum the first 249 integers starting from 25, so we can rewrite our sum using this little trick:[tex]\displaystyle \sum_{k=25}^{249}k = \sum_{k=1}^{249}k - \sum_{k=1}^{24}k[/tex]In other words, we're summing all the first 249 integers, but then we remove the first 24. As a result, we have the sum of all integers from 25 to 249:[tex]1+2+3+\ldots+248+249-(1+2+3+\ldots+23+24) = 25+26+27+\ldots+248+249[/tex]So, we have[tex]\displaystyle 4\sum_{k=25}^{249}k = 4\left(\sum_{k=1}^{249}k-\sum_{k=1}^{24}k\right)[/tex]And in both cases we can use the formula[tex]\displaystyle \sum_{k=1}^{N}k=\dfrac{N(N+1)}{2}[/tex]And we have[tex]\displaystyle 4\left(\sum_{k=1}^{249}k-\sum_{k=1}^{24}k\right)=4\left(\dfrac{249\cdot 250}{2}-\dfrac{24\cdot 25}{2}\right)=2\cdot 249\cdot 250 - 2\cdot 24\cdot 25[/tex]Which evaluates to[tex]2\cdot 249\cdot 250 - 2\cdot 24\cdot 25 = 124500-1200 = 123.300[/tex]