SOLUTION: a besiege fortress is held by 5700 men who have provision for 66 days. if the garrison losses 20 men each day, for how many days can the provision hold out?

Algebra ->  Sequences-and-series -> SOLUTION: a besiege fortress is held by 5700 men who have provision for 66 days. if the garrison losses 20 men each day, for how many days can the provision hold out?      Log On


   



Question 361865: a besiege fortress is held by 5700 men who have provision for 66 days. if the garrison losses 20 men each day, for how many days can the provision hold out?
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = amount of provision for each man for each day.
Then 5700x+%2B+5680x+%2B+5660x+... +%285700+%2B+%28n-1%29%2A%28-20%29%29x+=+66%2A5700x.
Dividing both sides by x, we get
5700+%2B+5680+%2B+5660+... +%285700+%2B+%28n-1%29%2A%28-20%29%29+=+66%2A5700,
5700+%2B+5680+%2B+5660+... +%285720-20n%29+=+66%2A5700.
The left side is the sum of an arithmetic sequence, whose sum is
S+=+%28n%2F2%29%2A%285700%2B5720+-+20n%29+=+n%285710-10n%29. Thus
n%285710-10n%29+=+66%2A5700, or
n%28571+-+n%29+=+66%2A570, or
n%5E2+-+571n%2B37620+=+0, or
%28n-495%29%28n-76%29+=+0.
Thus n = 495, or n = 76. The first value won't satisfy the 1st or the 2nd equation at the start of this solution, so n = 76. Therefore the garrison can hold out for 76 days.