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

1/2n(2a+(n-1)d)

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.

10000-n(3(n-1)+54)+3^(n-1)

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?

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  • $\begingroup$ 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
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If I understand the question correctly, you are being asked to compute

$$10000-54-57-\cdots-(51+3\cdot66)$$

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

$$(51+3\cdot1)+(51+3\cdot2)+\cdots+(51+3\cdot66)=51\cdot66+3(1+2+\cdots+66)$$

Can you take it from here?

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  • $\begingroup$ 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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