arithmetic sequence problem

**oblixps** · February 26th 2009, 03:39 PM

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.

**HallsofIvy** · February 26th 2009, 06:29 PM

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.

**arpitagarwal82** · February 26th 2009, 06:36 PM

nth term of sequence can be find by $S_n - S_{n-1}$
Simple

$a_n = (3n^2 - n ) - (3(n-1)^2 - (n-1))$
= $3n^2 - n - 3n^2 - 3 +6n + n -1$
= $6n -4$

**Soroban** · February 26th 2009, 07:24 PM

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: . $a_n \:=\:a + (n-1)d \:=\:2 + (n-1)6 \quad\Rightarrow\quad a_n \:=\:6n-4$

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