arithmetic sequences

[searchthestars9]

help! my teacher didnt explain arithmetic sequences very well, so im a bit confused.

how would you figure out something like

a=5, d= -3 : find the nth term, when the initial term and difference are given.

how do you just figure something like that out?

Comments

[fiberpunk]

I'm assuming you're saying that the initial term is 5 and the difference is -3. It's not obvious what you're using a and d for.

We think about this like robots. We calculate the first few terms:

T_1 = 5
T_2 = T_1 + -3 = 5 + -3
T_3 = T_2 + -3 = (5 + -3) + -3
T_4 = T_3 + -3 = ((5 + -3) + -3) + -3
etc.

Well, already I see a pattern. In the second term, I'd added one -3. In the fourth term, I'd added three -3s. I bet that in the n'th term, I will have added (n-1) -3s. In other words, I conjecture that:

T_n = 5 + (n-1)(-3)

Proof is via an easy induction.

[magdaz]

^yes, for the proof, the thoughtful method... But if you're looking to just quickly use the formula, figure it out using this:

--> a + (n-1)d
5 + (n-1)-3
5 + 3n + 3

3n + 8 --If that's what you mean...? Unless they expect the proof for this formula, then yeah, as posted above. That's the best way to understand how it works.

It's later incorporated in finding the sum of terms in a sequence:

Sn = 1/2*n [2a + (n-1)d]

So... anyone care to find proof for this formula? I would very much like to see a derived proof.

[fiberpunk]

The formula for the sum can be thought of this way: the n'th term has a few components, the a and (n - 1) ds. So to sum the first n terms, you are going to need n as, and then you're going to have to sum up (0 ... (n-1)) ds. That's, then, na + [sum of (0 ... (n-1)] * d.

The sum of the first (n-1) integers is n(n-1)/2 (this is a common formula, of which there are many proofs). So the sum should be:

na + [n(n-1)d]/2, and we take out an n, giving us:

n(a + [(n-1)d]/2), and we add the fractions so as to factor out the 1/2, getting us:

(n/2)[2a + (n-1)d]

[magdaz]

very niiice.

Yeah, that makes sense. Thanks.