Understanding Arithmetic Sequences and Formulas

In today's video, we're going to cover how to write an expression for the nth term of an arithmetic sequence. And remember, an arithmetic sequence is just a string of numbers where the same value is being added or subtracted each time. So if we look at this sequence here of 5, 9, 13, 17, and 21, it counts as an arithmetic sequence because it's increasing by 4 each time. We can represent sequences like this with expressions, and in this case the expression would be 4n plus 1. And the main benefit of doing this is that these expressions allow us to find any term in the sequence. If you're not sure what I mean by the word term, each of the numbers in the sequence is known as a term. So 5 is the first term, 9 is the second term, 13 is the third term, and so on. Another way to represent the terms is using the letter n. So for the first term, n equals 1, for the second term, n equals 2, and so on. So if we were asked to find the third term in the sequence, we're looking for the term where n equals 3. So all we have to do to find that is plug the number 3 into our expression. So we'd do 4 times 3, because n is 3, plus 1, which gives us 12 plus 1, or 13. So we know that the third term is 13. Now obviously for the third term we could have just looked at our sequence and seen that it's 13. But for bigger terms, like say the 50th term, we can't just look at our sequence, and it would take us ages to keep on adding 4 until we get to the 50th term. So instead, because we know that n will be 50, we can plug 50 into the expression to get 4 times 50 plus 1, which is 200 plus 1, so the 50th term is 201. Now the slightly trickier part is how to figure out the expression for a sequence in the first place. For this though we can use a neat First we need to find the common difference of our sequence, which is the number that's being added each time. So in this sequence that would be 4, because 5 add 4 is 9, 9 add 4 is 13, and so on. Then we take this common difference and place it in front of the letter m, so our expression will start with 4n. Then to find out what's going to be added or subtracted, we have to imagine that there's a previous term before the first term. So because the common difference here is 4, and our first term at the moment is 5, the previous term must be 4 less than 5, so positive 1. And then we include this term in our expression. So here that means that we put the plus 1 onto our 4n to get 4n plus 1 as our expression. So to summarize this, Because this sequence increases by 4 each time, we put 4n, and then because the imaginary term before the first term would be positive 1, we put plus 1 on the end to get 4n plus 1. Let's use this same technique to try and find the expressions for these two sequences as well. In our first sequence, the terms are decreasing by 5 each time. So the common difference is negative 5, and this means that our expression will start with minus 5n. Then to find the imaginary first term, we need to do the opposite and add 5 to the 26, which will give us 31. So we add 31 to our expression to get minus 5n plus 31. or if you wanted to you could write it as 31 minus 5n. It's completely up to you which way around you want to write it, but we do normally put the n term first. For this next one the common difference is positive 1.5 because it's increasing by 1.5 each time. So we start our expression with 1.5n. Then if we go back by one term and subtract 1.5 from the 1. we get negative 0.5. So we have to subtract 0.5 from our 1.5n to give us the expression 1.5n minus 0.5. Now just before we finish, a good trick to double check if your expression is correct or not is to try it for one of the terms that you already know. For example, we could try using 1.5n minus 0.5. to find the fifth term in our sequence, which we know should be 7. So we do 1.5 times 5, because remember n is 5 for the fifth term, and then take away the 0.5. So 7.5 minus 0.5, which indeed does give us 7. So we can be pretty confident that our expression is correct. Anyway, That's everything for this video, so if you found it useful, then please do tell your friends and your teachers about us, and thanks for watching!

We can represent sequences like this with expressions, and in this case the expression would be 4n plus 1. And the main benefit of doing this is that these expressions allow us to find any term in the sequence. If you're not sure what I mean by the word term, each of the numbers in the sequence is known as a term. So 5 is the first term, 9 is the second term, 13 is the third term, and so on. Another way to represent the terms is using the letter n. So for the first term, n equals 1, for the second term, n equals 2, and so on.

So if we were asked to find the third term in the sequence, we're looking for the term where n equals 3. So all we have to do to find that is plug the number 3 into our expression. So we'd do 4 times 3, because n is 3, plus 1, which gives us 12 plus 1, or 13. So we know that the third term is 13. Now obviously for the third term we could have just looked at our sequence and seen that it's 13. But for bigger terms, like say the 50th term, we can't just look at our sequence, and it would take us ages to keep on adding 4 until we get to the 50th term. So instead, because we know that n will be 50, we can plug 50 into the expression to get 4 times 50 plus 1, which is 200 plus 1, so the 50th term is 201. Now the slightly trickier part is how to figure out the expression for a sequence in the first place.

For this though we can use a neat First we need to find the common difference of our sequence, which is the number that's being added each time. So in this sequence that would be 4, because 5 add 4 is 9, 9 add 4 is 13, and so on. Then we take this common difference and place it in front of the letter m, so our expression will start with 4n. Then to find out what's going to be added or subtracted, we have to imagine that there's a previous term before the first term.

So because the common difference here is 4, and our first term at the moment is 5, the previous term must be 4 less than 5, so positive 1. And then we include this term in our expression. So here that means that we put the plus 1 onto our 4n to get 4n plus 1 as our expression. So to summarize this, Because this sequence increases by 4 each time, we put 4n, and then because the imaginary term before the first term would be positive 1, we put plus 1 on the end to get 4n plus 1. Let's use this same technique to try and find the expressions for these two sequences as well.

In our first sequence, the terms are decreasing by 5 each time. So the common difference is negative 5, and this means that our expression will start with minus 5n. Then to find the imaginary first term, we need to do the opposite and add 5 to the 26, which will give us 31. So we add 31 to our expression to get minus 5n plus 31. or if you wanted to you could write it as 31 minus 5n. It's completely up to you which way around you want to write it, but we do normally put the n term first. For this next one the common difference is positive 1.5 because it's increasing by 1.5 each time.

So we start our expression with 1.5n. Then if we go back by one term and subtract 1.5 from the 1. we get negative 0.5. So we have to subtract 0.5 from our 1.5n to give us the expression 1.5n minus 0.5. Now just before we finish, a good trick to double check if your expression is correct or not is to try it for one of the terms that you already know.

For example, we could try using 1.5n minus 0.5. to find the fifth term in our sequence, which we know should be 7. So we do 1.5 times 5, because remember n is 5 for the fifth term, and then take away the 0.5. So 7.5 minus 0.5, which indeed does give us 7. So we can be pretty confident that our expression is correct. Anyway, That's everything for this video, so if you found it useful, then please do tell your friends and your teachers about us, and thanks for watching!

Transcript for:Understanding Arithmetic Sequences and Formulas

Transcript for:
Understanding Arithmetic Sequences and Formulas