Question

How do you find the nth term rule for `34,25,16,7,-2,...`?

Answer 1

If you consider `a_0=34`, then `a_n = a_0-9n`
If you consider `a_1=34`, then `a_n = a_0-9(n-1)`

Explanation:

First of all, we need to see that the rule of the sequence is "subtract nine", and I suppose that this is the easy part.

Now, we need to see the sequence with the point of view of the first term: we start with `34`, and obtain `25` by subtracting `9`. Now, what's the link between the first and the third element? We obtain `16` by subtracting `9` again from `25`, which means that we are subtracting `9` twice from `34`.

This should lead you to the general rule: start with a number, subtract `9` once to obtain the second number, subtract `9` twice to obtain the third number...and so on.

Using a proper notation, if we define `a_0` as the first term of the sequence, we can write

`a_1 = a_0-9`
`a_2 = a_1-9 = (a_0-9)-9 = a_0-2*9`
`a_3 = a_2-9 = ( a_0-2*9)-9 = a_0-3*9`
`\vdots`
`a_n = a_0 - n*9`

which would be your closed formula for `a_n`

NB: if you choose to start with `a_1` instead of `a_0`, then you must rescale your coefficients, and you would have `a_n = a_1 - (n-1)*9`