Find the three consecutive odd numbers whose sum is 147.

Answer 1

a1 = first number

a2 = second number

a3 = third number

a2 = a1 + 2

a3 = a2 + 2 = a1 + 2 + 2 = a1 + 4

a1 + a2 + a3 =

a1 + a1 + 2 + a1 + 4 =

3 a1 + 6 = 147 Subtract 6 to both sides

3 a1 + 6 - 6 = 147 - 6

3 a1 = 141 Divide both sides by 3

a1 = 141 / 3 = 47

a1 = 47

a2 = 47 + 2 = 49

a3 = 49 + 2 = 51

47 + 49 + 51 = 147

Answer 2

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