Find the three consecutive integers that the sum of twice the smallest and 3 times the largest is 126.

Answer 1

Let x, x+1, and x+2 be three consecutive integers

2x + 3(x+2) = 126

2x + 3x + 6 = 126

5x = 120

x = 120 ÷ 5 = 24

The numbers are 24, 25 and 26.