Question

find four consecutive odd integers such that the sum of twice the smallest and the largest is 11 more than the sum of the middle two integers

Answer 1

x+(2+x)+(4+x)+(6+x)= x+12

2x + 6+x = 3x +6

2+ x + 4+x= 2x + 6

2x + 6 + 11 = 3x+6

2x + 17 = 3x +6

3x-2x = 17-6

x = 11

therefore the four consecutive odd integers are

11, 13, 15 and 17