Answer:
The four consecutive multiples of three are 6, 9, 12, and 15
Step-by-step explanation:
let the first multiple of 3 = n
second multiple of 3 = (n + 3)
third multiple of 3 =(n + 6)
fourth multiple of 3 =(n + 9)
product of the first and the third = n(n + 6) = n² + 6n
eight times the fourth number = 8(n + 9) = 8n + 72
The difference between the two = n² + 6n - (8n + 72) = -48
n² + 6n - 8n - 72 = -48
n² - 2n = -48 + 72
n² - 2n = 24
n² - 2n - 24 = 0
n² - 2n - 24 = 0
n² + 4n - 6n - 24 = 0
n(n + 4) - 6(n + 4) = 0
(n - 6)(n + 4) = 0
n = 6 or - 4
n should be a positive number, thus n = 6
The second multiple of 3 = (n + 3) = 6 + 3 = 9
The third multiple of 3 =(n + 6) = 6 + 6 = 12
The fourth multiple of 3 = (n + 9) = 6 + 9 = 15
