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?

score 1 · Answer 1 · answered Nov 07 '16 at 11:51

1

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

answered Nov 07 '16 at 11:51

lab bhattacharjee

274,582

Thank you, but as I said I know how to use differentiation for this question but I wanted another way. – oshhh Nov 07 '16 at 13:38

score 1 · Accepted Answer · answered Nov 07 '16 at 11:51

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} $.

answered Nov 07 '16 at 11:51

RGS

9,719

Oh right, how dumb of me to not think of that! – oshhh Nov 07 '16 at 13:39
@OsheenSachdev sometimes we overestimate a problem! We must always try to think in the simpler terms possible! – RGS Nov 07 '16 at 13:53

Finding the maximum sum of a decreasing Arithmetic Progression

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