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