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?
Best Answer
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?