I realise how to find the sum up a finite arithmetic series when the common ratio is the same each time.


However what happens when d (the common ratio) changes each time?

EDIT: I do not want to give too many details because it is an assignment and I'm trying to understand it rather than get given the answer.

I am given the first term to be 10000. I am then told when n = 1, an amount of 54 is taken off. then for each subsequent term an amount is removed from the previous value that is 3 more. I need to find how much is left of the original 10000 after 66 terms.

Now if the difference varies I am unsure of the formula to calculate the answer. Any pointers in the right direction would be much appreciated. Even if its just to a website that details the working.

EDIT2: After the comment about the sum of the differences I tried to think of an equation to calculate the sum of a given term but couldn't do it.


This worked great up until the fourth term... Sorry I'm not that good at Maths and have to work really hard to get things.

EDIT3: Another different approach. For reference there are 66 terms with term 0 being 10000.

1st Term = 54
Last Term = 54+3(n-1) = 54+3(66-1) = 249

So the sum of the accumulated negatives.

= 1/2 * (number of terms) * (first term + last term)
= 1/2 * 66 * (54 + 249)
= 9999

Therefore the amount left after 66 terms is as follows,

10000 - 9999 = 1

Any good or not?

  • $\begingroup$ Then it's not an arithmetic sereis. $\endgroup$ – Adam Hughes Jul 14 '14 at 20:56
  • $\begingroup$ 1) Arithmetic series gave common differences. Series with common ratios are called geomtric series. 2) The series you gave is neither. $\endgroup$ – David H Jul 14 '14 at 20:56
  • $\begingroup$ Added more detail to my question. I didn't originally because I don't want to cheat $\endgroup$ – Disco S2 Jul 14 '14 at 21:04
  • $\begingroup$ Hint: Let us concentrate on the total amount removed. If I understand the problem, the amount removed after say $5$ cycles is $54+(54+3)+(54+6)+(54+9)+(54+12)$. (I will say no more.) $\endgroup$ – André Nicolas Jul 14 '14 at 21:13
  • $\begingroup$ I'm sorry I'm still not getting any closer. Please see answer $\endgroup$ – Disco S2 Jul 14 '14 at 21:42

If I understand the question correctly, you are being asked to compute


Note that there are $66$ terms being subtracted: $54=51+3\cdot1$, $57=51+3\cdot2$, etc. So what's being subtracted from $10000$ is


Can you take it from here?

  • $\begingroup$ please see original question for attempted answer. I'm sorry I'm frustratingly slow $\endgroup$ – Disco S2 Jul 14 '14 at 22:14
  • $\begingroup$ @DiscoS2, it looks to me from your edit that you got it. $\endgroup$ – Barry Cipra Jul 14 '14 at 22:19
  • $\begingroup$ I like maths, I just don't have a gift for it. Thanks for your help. $\endgroup$ – Disco S2 Jul 14 '14 at 22:25

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