Question

Find 4 consecutive odd integers such that 5 times the sum of the first two was 10 less than 7 times the sum of the second and fourth. What are the four integers?

When I worked the problem, I got -13 as N and then when I finished it, it didn't seem correct. Can anyone help me?

Answer 1

let the smallest of the odd integers be x
then the next 3 consecutive integers are
x+2,x+4, and x+6

5(x + x+2) = 7(x+2 + x+6) - 10
10x + 10 = 14x + 46
-4x = 36
x = -9

so the 4 integers are -9, -7, -5, and -3

check:
5times(sum of first two) = 5(-9 -7) = -80
7times(sum of 2nd and 4th) = 7(-7 -3) = -70

Is -80 ten less than -70? YES