# Arithmetic sequence problem

1. Feb 25, 2009

### demonelite123

The sum of the first n terms in a certain arithmetic sequence is given by Sn = 3n2 - n. Show that the nth term of the sequence is given by an = 6n - 4.

so far i have done:
Sn = (n / 2) (a1 + an) = 3n2 - n
i solved for a1 + an = 6n - 2

i also have an = a1 + d(n-1).

2. Feb 25, 2009

### mjsd

d = 6, from $${d'}_{n} = S_{n+1} - S_{n}$$ and then evaluate $$d = {d'}_{n+1} - {d'}_n =6$$
and note that $$a_1 = S_1$$

3. Feb 25, 2009

### demonelite123

i don't understand what dn is. isn't d just the common difference? how come there's an apostrophe on it?

4. Feb 26, 2009

### mjsd

the way I have written it, $$d'$$ is not actually $$d$$ in the defintion of: $$a_n = a_1 + d (n-1)$$
so I put the "prime" or apostrophe on it. But the difference between two consecutive $$d'$$ is the $$d$$ we are after... write down the sequence and the progression and work out the differences between consecutive entries to see the pattern and visualise how these results are derived.

5. Feb 26, 2009

### demonelite123

sorry i don't understand this part: Sn+1 - Sn = d' and then evaluate d = d'n+1 - d'n = 6.

how did you find what d'n+1 was? and how did you know d = 6?

6. Feb 26, 2009

### HallsofIvy

Staff Emeritus
You are told that the sum of the first n terms is 3n2- n. Then the first term, alone, a1= 3(12)- 1= 2. Also the sum of the first two terms is a1+ a2= 3(22)- 2= 10 so a2= 10- a1= 10- 2= 8. So the first term is 2 and the common difference is 8-2= 6. The nth term is 2+ 6(n-1)= 2+ 6n- 6= 6n- 4.

7. Feb 26, 2009

### demonelite123

thanks a bunch!!
i don't know why i didn't think of that!