# Thread: Arithmetic Sequence help!

1. ## Arithmetic Sequence help!

So I was going over some of my Algebra II work and the part that we're going over is COMPLETELY different than the practice problems. We have this question that I didn't even see the directions for how to solve like this:

What are the missing terms of the arithmetic sequence -8, ___, ___, 3

I know that the difference is 11 between the two numbers, but I've tried every way possible to get the correct answer. I don't understand how to do it at all. Help, please?

2. Originally Posted by delovely9109
So I was going over some of my Algebra II work and the part that we're going over is COMPLETELY different than the practice problems. We have this question that I didn't even see the directions for how to solve like this:

What are the missing terms of the arithmetic sequence -8, ___, ___, 3

I know that the difference is 11 between the two numbers, but I've tried every way possible to get the correct answer. I don't understand how to do it at all. Help, please?
$U_{n} = a + (n-1)d$

where $U_n$ is the value of term n, $a$ is the first term, $n$ is the number of terms and $d$ is the common difference.

In your case we have $a = -8$ and $U_4 = 3$

Putting what is known into the formula:

$3 = -8 + (4-1)d$ and it's not hard to see that $d = \frac{11}{3}$

Therefore your missing numbers are $-8 + \frac{11}{3} = -\frac{13}{3}$ and $-8+2\frac{11}{3} = -\frac{2}{3}$

3. Hello, delovely9109!

Another approach . . .

What are the missing terms of the arithmetic sequence: -8, ___, ___, 3
I must assume that you know something about Arithmetic Sequences.

We have an arithmetic sequence with common difference $d.$

. . $\begin{array}{cc}\text{The first term is}&\text{-}8 \\ \text{The second term is:}&\text{-}8+d \\ \text{The third term is:}&\text{-}8+2d \\ \text{The fourth term is:}&\text{-}8+3d \end{array}$

But we already know that the fourth term is $3.$

So we have: . $\text{-}8+3d \:=\:3\quad\Rightarrow\quad 3d \:=\:11\quad\Rightarrow\quad d \:=\:\tfrac{11}{3}$

Therefore, the sequence is: . $\text{-}8,\;\text{-}\tfrac{13}{3},\;\text{-}\tfrac{2}{3},\;3,\;\hdots$