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?