Finding the maximum sum of a decreasing Arithmetic Progression

Question

Find the maximum sum of the AP $40+38+36+34+32+...$

My Attempt:
$a=40$
$d=(-2)$
$S_n= \frac{n}{2}[2a+(n-1)d]$
$S_n= \frac{n}{2}[80+(n-1)(-2)]$
$S_n= n[40+(n-1)(-1)]$
$S_n= n[41-n]$
$S_n= 41n-n^2$

I can find the answer by differentiating the expression and finding the maxima but is there any other way obtaining the answer?

Answer 1

Sure there is. If you think for a second you understand that as soon as your progression hits 0, it will start adding negative numbers therefore decreasing its sum. So the maximum sum is $S_{20} $.

Answer 2

As $n$ integer, find when $$\dfrac{d(41n-n^2)}{dn}>=<0$$