Find three consecutive odd integers where four more than 3 times the first number plus the largest is 76.

Answer 1

Let x be the first number, then x +2, and x + 4 are the second and third numbers. Four more than three times the first number is 3x + 4

Adding that to the last number gives you (3x + 4) + (x + 4) = 76

4x + 8 = 76

4x = 68

x = 17

x + 2 = 19

x + 4 = 21

Answer 2

denote the odd sequence by (2n-1, 2n+1, 2n+3). We have:

4 + 3(2n-1) + (2n+3) = 76 --> 4 + 6n - 3 + 2n +3 = 4+ 8n=76 or n=9. Therefore the numbers are:

(17, 19, 21).