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Mathematics 19 Online
OpenStudy (anonymous):

The sum of 5 consecutive even numbers is 220. What is the fourth number in this sequence?

OpenStudy (xapproachesinfinity):

5 consecutive number let's right them like this n+(n+1)+(n+2)+(n+3)+(n+4)=220 so 5n+10=220 solve for n

OpenStudy (cwrw238):

let numbers be x , x+1, x+2, x+3, x+4 then x + x+1 + x+2 + x+3+ x + 4 = 220 find x you require the value of x + 3

OpenStudy (xapproachesinfinity):

oh hold! i didn't pay attention to even

OpenStudy (xapproachesinfinity):

let's call the first number 2k then we have 2k+ 2(k+1)+2(k+2)+2(k+3)+2(k+4)=220 solve for k

OpenStudy (cwrw238):

oops sorry i misread the question even numbers replace 1, 2, 3 and 4 by 2, 4 , 6 and 8 you require the value of x + 6

OpenStudy (anonymous):

@iGreen

OpenStudy (xapproachesinfinity):

you have 10k+20=220 k=200/10 k=20 the first number is 40, 42, 44, 46, 48 if we add all this number together we get 220 then the fourth number is 46

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