# arithmetic sequence problem

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• Feb 26th 2009, 03:39 PM
oblixps
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.
• Feb 26th 2009, 06:29 PM
HallsofIvy
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.
• Feb 26th 2009, 06:36 PM
arpitagarwal82
nth term of sequence can be find by $\displaystyle S_n - S_{n-1}$
Simple :)
$\displaystyle a_n = (3n^2 - n ) - (3(n-1)^2 - (n-1))$
= $\displaystyle 3n^2 - n - 3n^2 - 3 +6n + n -1$
= $\displaystyle 6n -4$
• Feb 26th 2009, 07:24 PM
Soroban
Hello, oblixps!

Yet another approach . . .

Quote:

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

Show that the $\displaystyle n^{th}$ term of the sequence is given by: $\displaystyle a_n \:= \:6n - 4$

The sum of the first $\displaystyle n$ terms of an arithmetic sequence: .$\displaystyle S_n \:=\:\frac{n}{2}\bigg[2a + (n-1)d\bigg]$

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

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

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

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

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

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

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