Nth term
From I2008
(Difference between revisions)
Revision as of 15:11, 22 June 2006
Let d = common difference n = position in sequence [stays as "n" in the formula] k = an addend dn + k = [value of term]
Contents |
Common Difference (d)
The common difference, d, is easy to find out.
For example:
In 6, 13, 20, 27, 34, 41, 48...
The common difference is 7.
Position in Sequence (n)
In 5, 7, 9, 11, 13, 15...
11 is in Position 4. This means that on that case, n=4.
If n=4 then the term=11. Sounds good?
Kehy (k)
k is an addend.
A sequence like 3, 6, 9, 12, 15 is pretty obvious.
What makes it more challenging is when you give it a twist.
You can add -4 to every term and get -1, 2, 5, 8, 11. You can add 150 to every term and get 153, 156, 159, 162, 115.
The possibilities are endless. This is what k does.
How to Get the Formula
dn + k = ? ___________________________________________ | TERM | 1 | 2 | 3 | 4 | 5 | 6 | | VALUE | 5 | 11 | 17 | 23 | 29 | 35 | '-------------------------------------------'
- Figure out the common difference
- In this case, the common difference is 6.
- So, we can conclude that the equation must contain: 6n + k = ?
- Now, how do we get k?
- To get k, substitute n with any given. For our purpose, let's use 3.
- We will substitute n with 3 and the question mark with the value of the third term..
- 6(3) + k = 17
- 18 + k = 17
- k = -1
- And so, we can now tell that k = -1.
- Bearing this in mind, we can now say that the formula for this specific linear function is: 6n - 1.