SOLUTION: what is the sum of the first 100 natural Numbers excluding the multiples if 4

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Question 1034674: what is the sum of the first 100 natural Numbers excluding the multiples if 4
Found 2 solutions by sachi, Edwin McCravy:
Answer by sachi(548) (Show Source):
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the sum first 100 natural Numbers i.e 1+2+3+.....+100
as per AP
the first term=1=a
the common diff =1-d
no of terms = 100=n
now the sum=n/2[2a+(n-1)*d)=100*101/2=5050
now the sum of multiples of 4=4+8+12+.....+100
the last term=100=l
the first term=4=a
the common diff =4-d
no of terms=n
then 100=a+(n-1)*d=4+(n-1)*4=4n
so n=25
now the sum of multiples of 4=4+8+12+.....+100=25/2(a+l)=25/2*(4+100)=1300
so the sum of the first 100 natural Numbers excluding the multiples if 4
=5050-1300=3750
ans

Answer by Edwin McCravy(20056) (Show Source):
You can put this solution on YOUR website!

That's (1+2+3+...+99+100)) - (4+8+12+...96+100) There are 100 natural numbers and there are 25 multiples of 4 through 100, because if you divide all the numbers 4,8,12,...,96,100 by 4, you get 1,2,3,...,24,25. We use the formula for the sum of an arithmetic series with n terms, first term a₁, nth term a_n. (1+2+3+...+99+100)) - (4+8+12+...96+100) = Edwin