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# tricky maths arithmetic question; maximum term watch

1. The information I have

First term = 43. Common difference =-3

15 positive terms.

"find the maximum value of sn. the sum of the first n terms of the series"

2. (Original post by ZiggyStarDust_)
The information I have

First term = 43. Common difference =-3

15 positive terms.

"find the maximum value of sn. the sum of the first n terms of the series"

Can you post some working? Have you found yet?
3. (Original post by Mr M)
Can you post some working? Have you found yet?
the maximum value of sn is what i'm trying to find.

with the information i've said before, is all that i'm given to work this out.

i've tried to use the formula: 0.5n(2a+(n-1)d)

But the value of n is not given in the question. so i am confused as to how to solve this question.
4. (Original post by ZiggyStarDust_)
the maximum value of sn is what i'm trying to find.

with the information i've said before, is all that i'm given to work this out.

i've tried to use the formula: 0.5n(2a+(n-1)d)

But the value of n is not given in the question. so i am confused as to how to solve this question.
I know what you are trying to find.

You do know the value of n because you provided it in the question.

If you didn't know it, you could differentiate to find the maximum but that would be over-the-top in my opinion.
5. When will a sum start to decrease in value?
6. (Original post by an_atheist)
When will a sum start to decrease in value?
43,40,37,34,31,28,25,22,19,16,13 ,10,,4,1

at the 15th term.
7. (Original post by ZiggyStarDust_)
The information I have

First term = 43. Common difference =-3

15 positive terms.

"find the maximum value of sn. the sum of the first n terms of the series"

(Original post by ZiggyStarDust_)
the maximum value of sn is what i'm trying to find.

with the information i've said before, is all that i'm given to work this out.

i've tried to use the formula: 0.5n(2a+(n-1)d)

But the value of n is not given in the question. so i am confused as to how to solve this question.

8. (Original post by ZiggyStarDust_)
43,40,37,34,31,28,25,22,19,16,13 ,10,,4,1

at the 15th term.
The maximum value for the summation of the first n terms of this series will therefore be?
9. (Original post by Mr M)
I know what you are trying to find.

You do know the value of n because you provided it in the question.

If you didn't know it, you could differentiate to find the maximum but that would be over-the-top in my opinion.
so 15 is N?
10. (Original post by ZiggyStarDust_)
so 15 is N?
Yes because if you start to add negative terms to your total it will reduce the total.
11. (Original post by will'o'wisp)

thanks very much, d*ckhead. you've been a great help to my problems.

go troll the chat side of TSR; you know, where no one studies/gives a hoot.

i'm actually trying to solve a bloody question here, if you don't know how to help, don't post anything, moron.
12. (Original post by Mr M)
Yes because if you start to add negative terms to your total it will reduce the total.
well, i tried putting 15 into the formula, but it didn't give me 330. (which should be the answer, according to the mark scheme)
13. (Original post by ZiggyStarDust_)
well, i tried putting 15 into the formula, but it didn't give me 330. (which should be the answer, according to the mark scheme)
What did you use as a and d?
14. (Original post by Mr M)
What did you use as a and d?
a= 43

d=-3

0.5n(2a+(n-1)d) - formula used
15. (Original post by ZiggyStarDust_)
a= 43

d=-3

0.5n(2a+(n-1)d) - formula used
Looks like you are failing to operate your calculator properly.

Try typing this in exactly.

16. (Original post by Mr M)
Looks like you are failing to operate your calculator properly.

Try typing this in exactly.

that was 330! So all this time, i was doing nothing wrong, i just mistyped something in my calculator

thank you very much for your help, Mr M. You really helped me out there, it's very much appreciated. you rock, my dude. you rock
17. (Original post by ZiggyStarDust_)
thank you very much for your help, Mr M.
You are welcome.

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