Finding the nth Term of an Arithmetic Sequence

[demonelite123]

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

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

i also have a_n = a₁ + d(n-1).

i don't know what do to next. please help me.

 

[mjsd]

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

 

[demonelite123]

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

 

[mjsd]

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

the way I have written it, $$d'$$ is not actually $$d$$ in the definition 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.

 

[demonelite123]

sorry i don't understand this part: S_n+1 - S_n = 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?

 

[HallsofIvy]

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.

 

[demonelite123]

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