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
Find the three consecutive odd numbers whose sum is 147.
2 answers
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Looks like you want us to do your homework or your assignment for you.
I did one of them for you, show me some work or effort on your part for the others, and I will help if you run into difficulties.