Question

Find three consecutive odd integers such that four times the middle integer is equal to two less than the sum of the first and third integers.

Answer 1

The three consecutive odd integers are -3, -1, and 1. This satisfies the condition: four times the middle integer equals two less than the sum of the first and third integers.

Let's represent the three consecutive odd integers as n, n+2, and
n+4 , where n is the first odd integer. According to the given conditions:

1. Four times the middle integer: 4(n+2)

2. Sum of the first and third integers: (n) + (n+4)

3. Two less than the sum: (n) + (n+4) - 2

Now, we set up the equation:

4(n+2) = (n) + (n+4) - 2

Solving for n :

4n + 8 = 2n + 2

Combine like terms:

2n = -6

Divide by 2:

n = -3

So, the three consecutive odd integers are -3, -1, and 1 .