# Math Help - arithmetic sequence problem

1. ## arithmetic sequence problem

sry it's me with another question again. my test is coming up and my teacher is giving us all these challenge problems to prepare.

The sum of the first n terms in a certian arithmetic sequence is given by Sn = 3n^2 - n. Show that the nth term of the sequence is given by an = 6n - 4.

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

i also have the equation a_n = a1 + d(n-1)

i have tried solving for a_n, a1, and d using these equations but i'm not getting anywhere. Please help me.

2. A term in an arithmetic sequence can always be written a+ bn so you only need to find two values, a and b. For that you need two equations. Apply what you know about "the sum of the first n terms" to the sum of the first term only and the sum of the first two terms.

3. nth term of sequence can be find by $S_n - S_{n-1}$
Simple
$a_n = (3n^2 - n ) - (3(n-1)^2 - (n-1))$
= $3n^2 - n - 3n^2 - 3 +6n + n -1$
= $6n -4$

4. Hello, oblixps!

Yet another approach . . .

The sum of the first $n$ terms of an arithmetic sequence is given by: $S_n \:=\: 3n^2 - n$

Show that the $n^{th}$ term of the sequence is given by: $a_n \:= \:6n - 4$
The sum of the first $n$ terms of an arithmetic sequence: . $S_n \:=\:\frac{n}{2}\bigg[2a + (n-1)d\bigg]$

We will take $S_n \:=\:3n^2-n$ and hammer it into that form . . .

We have: . $S_n \:=\:n(3n-1)$

Multiply by $\tfrac{2}{2}\!:\;\;\frac{2}{2}\cdot n(3n-1) \:=\:\frac{n}{2}(6n-2)$

Subtract 6 and add 6: . $\frac{n}{2}\bigg[6n {\color{blue}\:-\: 6\: +\: 6} - 2\bigg] \:=\;\frac{n}{2}\bigg[6(n-1) + 4\bigg] \;=\;\frac{n}{2}\bigg[4 + (n-1)6\bigg]$

So we have: . $S_n \;=\;\frac{n}{2}\bigg[2(2) + (n-1)6\bigg]$

Hence, the first term is $a = 2$ and the common difference is $d = 6$

Therefore, the $n^{th}$ term is: . $a_n \:=\:a + (n-1)d \:=\:2 + (n-1)6 \quad\Rightarrow\quad a_n \:=\:6n-4$