# Arithmetic sequence problem

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).

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

## Answers and Replies

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

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

mjsd
Homework Helper
i don't understand what dn 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 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.

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?

HallsofIvy