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]
Answer:
First 3 digit multiple = 100
Last 3 digit multiple = 996
AP: 100, 104, 108...996
an = 996
a + (n-1)d = 996
100 + (n-1)4 = 996
(n-1)4 = 896
(n-1) = 224
Therefore n = 225
Sn = n(a+l)/2
= 225(100+996)/2
= 225 x 548
= 123300
Hope this helps!
