Jim's Market couldn't keep Crunchy Critter Crackers in stock. Jim started with 300 boxes but everyone wanted them. The first day Jim sold 6 boxes, and on the second day he sold 14 boxes. Each day 8 more boxes were sold than the day before. So after two days, he had sold 20 boxes. If he kept selling the crackers at this rate, when would Jim run out of Crunchy Critter Crackers?
This is the sum of an arithmetic series
n ( a1 + a1 + (n -1) d) / 2 > 300
a1 = 6 = the first term
n = the number of days
d = the common difference = 8
So we have
n ( 6 + 6 + (n-1) * 8 ) / 2 > 300
n ( 12 + 8n - 8) > 600
n ( 4 + 8n) > 600
4n + 8n^2 > 600 divide through by 4
2n^2 + n > 150
2n^2 + n - 150 > 0
Solve this
2n^2 + n -150 = 0
Using the Quadratic Formula
n = -1 + sqrt ( 1^2 - 4 (2) (-150) ) -1 + sqrt ( 1201)
_________________________ = _______________ ≈ 8.41
2 * 2 4
He will run out sometime during the 9th day