SOLUTION: Find 3 consecutive numbers whose sum is 20 more than the second number

Question 773700: Find 3 consecutive numbers whose sum is 20 more than the second number
Found 2 solutions by ramkikk66, Menjax:
Answer by ramkikk66(644) (Show Source): You can put this solution on YOUR website!
Let the 3 numbers be n-1,n and n+1. The middle number is n.
Their sum = n-1+n+n+1 = 3*n.
Now from the problem statement
3*n = n + 20.
2*n = 20 or n = 10.
So the 3 numbers are 9,10 and 11.
:)

Answer by Menjax(62) (Show Source): You can put this solution on YOUR website!
x + (x+1) + (x+2) = 20 + (x+1)
3x+3=21+x
2x=18
x=9

the numbers are 9,10,11
9 + 10 + 11 = 20 + 10
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