Question

Find seven consecutive even integers such that the sum of the first and the last integer is equal to the opposite of the fourth integer increased by twice the value of the fifth integer.

Answer 1

To find seven consecutive even integers where the sum of the first and the last equals the opposite of the fourth plus twice the fifth, you assign variables and solve an equation. The integers are found to be 2, 4, 6, 8, 10, 12, and 14.

Step-by-step explanation:

The question here involves finding seven consecutive even integers where certain conditions apply. First, let's define the first integer as x. Therefore, the consecutive even integers can be expressed as x, x + 2, x + 4, x + 6, x + 8, x + 10, and x + 12. According to the problem statement, the sum of the first and the last integer (x + (x + 12) = 2x + 12) should be equal to the negative of the fourth integer plus twice the fifth integer (-(x + 6) + 2(x + 8)). This gives us the equation 2x + 12 = -x - 6 + 2x + 16. Simplifying this equation, we get x = 2. Therefore, the seven consecutive even integers are 2, 4, 6, 8, 10, 12, and 14. We can check our answer by verifying the conditions: 2 + 14 = 16, and the opposite of the fourth integer (8) increased by twice the fifth integer (20) is indeed 16.