# Thread: Finding the next four terms of this sequence?

1. ## Finding the next four terms of this sequence?

5n, -n, 7n...

2a-5, 2a+2, 2a+9...

I'm having trouble with the variables. Any problem solving help?

2. Originally Posted by puzzledwithpolynomials
3, 4.5, 6...
My brain is just not working today? I don't know. I was thinking to maybe use the sequence forumla and try to derive it from there?
Also:
5n, -n, 7n...
For the simplest n'th term sequence, the formula is $\mathrm{n^{\mathrm{th}} \ Term} = dn+(a-d)$ where $d$ is the difference between the terms, $a$ is the first initial term and $n$ is the term number. Thus for $3,4.5, 6$, the difference between the term ( $d$) is $+1.5$, the first term ( $a$) is $3$ thus the formula is $\mathrm{n^{\mathrm{th}} \ Term} = (1.5)n + (3-1.5) = 1.5n + 1.5$.

For the second, do you mean $5n, -n, -7n$? This would make the difference $-6n$, the first initial term $5n$ and thus you would be able to use the n'th term formula.

3. Originally Posted by puzzledwithpolynomials
5n, -n, 7n...

2a-5, 2a+2, 2a+9...

I'm having trouble with the variables. Any problem solving help?
Hi puzzled,

Are you sure about the first 3 terms of your first sequence? It would be nice if it were:

5n, -n, -7n

Then the common difference would be -6n. You simply add -6n four more times to get your four terms.

In the second one, 2a-5, 2a+2, 2a+9, you have a common difference of 7. Looks like each successive term is 7 more than the previous one. So, just add 7 four more times to get the next four terms.