# Video: Describing Relationships and Extending Terms in Arithmetic Sequences

Through examples, describe relationships in arithmetic sequences. Then extend the sequences to 𝑛 positions, with 𝑛th term formulae being a simple multiple of 𝑛 (e.g., 6𝑛, 51𝑛, etc.). Use tables to clearly see the relationships in sequences.

08:40

### Video Transcript

Describing Relationships and Extending Terms in Arithmetic Sequences. Here is an example of an arithmetic sequence. A sequence is any ordered list of numbers. An arithmetic sequence is a sequence found by adding the same number to the previous term. For example, you add nine to 28. And that equals 37. You add nine to 37. And that equals 46. Each of these numbers are terms in the sequence.

Someone might ask you, find the next term in this arithmetic sequence.

When they say, find the next term, we know that they’re looking for the value that comes after 64 here. And because we’re also given the information that it’s an arithmetic sequence, we also know that we will be adding to find the next value. Each step in this sequence has been adding nine to the previous term. This lets us know that whatever 64 plus nine is, that’s our next term. The next term in this sequence is 73.

Here’s an example of a simple arithmetic sequence. We’re adding two each time.

When we’re dealing with sequences, each of the terms also has a position. You can see the labels here. The two is in first position. The four is in second position. Six is in third position and then fourth and fifth, and so on. Positions are really important. Let’s take a closer look at what is happening here with these positions. Position one has a value of two. Position two has a value of four. To get from position one to position two, we need to add two. And to go from position two to position three, we need to add two. There is a relationship here between the position of a term and the term’s value. This column shows us that, to solve for position one, we multiply one times two. To solve for position two, the value, we take two. And we multiply it by two. To solve for position three, you multiply three by two to get six.

In the operation, the first number we’re looking at is the position number. The two is the number that we’re adding to each term. Some of you might be wondering, Well why would I take the time to figure out the multiplication when I could just add two to six? Six plus two is eight. That’s pretty simple. And that’s true. If I wanted to find the next number in this sequence, I would probably just say 10 plus two is twelve. The sixth position is twelve. But if the question said something like what is the 80th position in this sequence, you don’t wanna add two 80 times. Now we’re going to add the number 80, the position 80, to our table. We would follow the same operation. In the 80th position of this sequence is the number 160. Finding that by multiplying 80 times two is significantly faster than trying to add two each time 80 times.

You might also come across the question: what is an algebraic expression for finding terms in this sequence? In that case, we don’t know what position we’re looking for. We’re looking to write an expression that can be used for all positions. When dealing with positions and sequences, we usually use the letter 𝑛 to represent an unknown position. We could find the value of a term in position 𝑛 by multiplying the position by two, 𝑛 times two. We could say 𝑛 times two for our expression, or simply two 𝑛. We can plug in any position number here and find the term that would be in that position. We could plug in position 100, position seven, position 15. It doesn’t matter. This expression works for solving this sequence. Here’s another example.

Find the next term in this sequence. 51, 102, 153. What’s next?

The first question we should ask is what is being added to each term. Last time, that was really easy because it was two. If you don’t immediately recognise what’s being added, here’s what you can do. You can take the term in the second position and subtract the term in the first position. 102 minus 51 is 51. You could also subtract 102 from 53 or the position two number from the position three number. Both of these equal 51. 51 is what is being added each time.

To find the next term, we need to take the third term, 153, and add 51 to that. The next term in this sequence is 204. Here’s what a table for this sequence would look like. The operation here would be to take the position and multiply it by 51 because 51 is what we’re adding here. 51𝑛 would be the expression that we could use to take any position and find its value.

Here is our last example.

Find the 70th term in the following sequence. Six, 12, 18, 24.

We’re trying to answer the question what is being added to each term. And we know that here each term is six more than the previous term. But I’m not just trying to find out what are we adding to each term. I need to find out what expression can I use to find the 70th term. I notice the pattern is that you take the position. And you multiply it by six. So in order for me to find the 70th term, I need to multiply that by six. When I do that, I have a solution of 420. I also know that I can find any term in this sequence by taking the position 𝑛 and multiplying it by six.

That expression is important. That’s all for this video. Now you can go try some sequences on your own.