Word problem of Arithmetic Series This question is taken from  my text book, given in the Arithmetic Series exercise.
I don't how to find the days using Arithmetic series?
Q: A besieged fortress is held by 5700 men who have provisions for 60 days. If the garrison loses 20 men each day, for how many days will the  provision last?
Thanks
 A: @zonnie
Each guy eats $\frac{1}{342000}$ of the food per day. So we have the sum
$$\frac{\sum _{k=0}^d (5700-20 k)}{342000}=1$$
The sum in the numerator is easy and the whole thing becomes
$$-\frac{(d-570) (d+1)}{34200}= 1 $$
solving the quadratic we get
$$d = 66.99080479207322, d = 502.0091952079268 $$
The second answer is meaningless so the food should hold out for 66 full days and on the 67th day they are hungry. But since we called the first day 0 ( in the sum ) we should add 1 so they will last 67 days and on the 68th they are out.
A: There are food for $5700\times 60=342 000$ days for 1 man.
$1$-st day they'll eat $5700$ daily food-items ($a_1=5700$).
$2$-nd day  they'll eat $5680$ daily food-items ($a_2=5680$).
$n$-th day $...$ $a_n=5700-20(n-1)$, ($n\le 285$, to have non-negative values of $a_n$).
If denote $$S_n = \sum_{k=1}^n a_n,$$ then $\Rightarrow$ condition:
$$S_n\le 342000.$$
Sum of arithmetic progression is:
$$
S_n=\dfrac{a_1+a_n}{2}\cdot n = \dfrac{5700+5720-20n}{2}\cdot n = (5710-10n)n,
$$
so we get
$$
(5710-10n)n\le342000;
$$
$$
10n^2-5710n+342000\ge 0;
$$
for $n=67$ we have $S_n=337680<342000$,
for $n=68$ we have $S_n=342040>342000$.
A: I think the assumption being made is that each man eats a constant amount of food each day, call it $r$. The statement that they have provisions for $60$ days is supposed to be interpreted as saying that, if the number of men remained constant, they would last $60$ days at that rate. So they have $60\cdot5700\cdot r$ units of food.
If $x_n$ is the amount of men still alive on day $n$, then on that day they'll eat $rx_n$ units of food. Can you work out the formula for $x_n$ and use that to figure out when they'll have eaten the $60\cdot5700\cdot r$ units they started with?
