5 consecutive even integers is 110

Question

5 consecutive even integers is 110

Answer 1

if n is our first integer we know the next integer would be n+2 because even numbers are all 2 apart (ex: 2+2 =4) the next even integer.

So 5 consecutive can be written as n,n+2,n+4,n+6,n+8, and n+10

Then it can be written as

n+n+2+n+4+n+6+n+8=110

5n+20=110

5n= 90

n= 18

so

18,20,22,24,26

Answer 2

18, 20, 22, 24, 26 sum to 110

x + (x+2) + (x+4) + (x+6) + (x+8) = 5x + 20

5x+20 = 110

5x = 90

x =90/5 = 18

x+2 = 20

x+4 = 22

x+6 = 24

x+8 = 26