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
@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.
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?