SOLUTION: a besieged fortress is held by 5700 men who have provisions for 66 days. of the garrison loses 20 men each day, for how many days will be provisions last?

There are 5700*66 portions of provision in the storage.


In the 1st day, 5700 portions are consumed.

In the 2nd day, 5700-20 = 5680 portions are consumed.

In the 3nd day, 5680-20 = 5660 portions are consumed.

. . . and so on . . . 


The problem wants you find the sum of the first n terms of this arithmetic progression

     = 


and the number n in such a way that  = 5700*66, under an additional condition  >= 0.


The first term of this AP is 5700; the common difference is -20,  so the sum of the first n terms is

     =  =  = .


So we write this equation

     = 5700*66


Simplify 

    20n^2 - 11420n + 11400*66 = 0

      n^2 - 571 + 5700*66 = 0.


Next, apply the quadratic formula.  You will get

       = .


The value  =  = 76 works: it satisfies  =  = 5700*66 and  >= 0.


The other value  =  = 495 does not work:  = 5700-20*495 = -4200 is negative.


So, the problem is just solved, and the  ANSWER  is 76 days.

What a gruesome problem! (Men getting killed in war!)

Let K = the number of units of provisions (k-rations] required for 5700 
men for 66 days.

Then K/2700 = the number of units of provisions required for 1 man for 
66 days.
 
Then K/(2700•66) = the number of units of provisions required for 1 man 
for 1 day.

If nobody were killed, the sequence over 66 days would be this
obvious sum of 66 terms with all identical terms:



However, 20 men are killed each day, so the sequence has more terms and
goes like this:



We want to find out how many terms the sequence must have to make 
the above equation true. 

We can divide both sides by K



We can factor out 



Suppose the sequence has N terms, then its sum is 

       , where a1=2700, and d=-20



Factor 20 out of the last parentheses on the left



Further simplifying:













N-76 = 0;     N-495 = 0
   N = 76;        N = 495

495 is extraneous.

Answer 76 days.

Edwin