## Arithmetic Sequences

### Transcript

Arithmetic sequences. One special kind of sequence is called an arithmetic sequence. An arithmetic sequence is one in which we add the same constant to get from each term to the next. So if you look at this sequence, you'll notice that to get from each term to the next, we're simply adding seven.

That's how we go from one term to the next. So we could continue this pattern forever, just continually adding seven to get the next number. Notice first of all, a really big idea. Any evenly spaced list is an arithmetic sequence.

So this is powerful. One special case is consecutive multiples of an integer p. We would add p to get from each term to the next. Consecutive odd numbers, consecutive even numbers, or consecutive integers are also special cases of arithmetic sequences. Here's an arithmetic sequence.

It would be helpful to find a formula for the nth term, so that, for example, we could find the value of a much higher term in the sequence. The test likes to ask about this. Remember for the algebraically defined sequences that we saw in the previous video, it's very easy. If the test said find the 80th term in this sequence, all we'd have to do is plug in.

But we'd like a formula that would allow us to do that for arithmetic sequences. I'm gonna analyze the individual terms in the following way to look for a pattern. So I'll focus on this particular sequence. The sequence that starts with five, and then we add seven each time. So the first term is five.

The second term is 12. I'm gonna write that as five plus seven. We've added seven once. To get the third term, we add seven again. So I'm gonna write that as five plus seven plus seven. Then the fourth term, we add seven again, then add seven again.

And so what we have here, this stream of sevens. We have five plus a number of sevens. Notice that the number of sevens, in each row, the number of sevens that we're adding is one less than the index number. It's one less than the number on the list. In each term the number of seven's added as one less than the index number of the term.

This is a big idea. Thus, the nth term, a sub n, would be the starting term plus n minus one individual sevens. Of course, a compact way to add n minus 1 individual 7's would simply be to multiply 7 times n minus 1. So that means that the nth term of this sequence would be 5 plus 7 times n minus 1.

Now we can generalize a bit. First of all, the starting term in general's a sub 1. We're familiar with that from the last video. The fixed amount we add each time to get each term is called the common difference. Because if we subtract any two adjacent terms, we get this same number.

We symbolize the common difference with the letter D. The nth term of an arithmetic sequence with an initial term A sub one and a common difference D, is A sub N equals A sub one plus N minus one times D. That is a very powerful formula. Please do not simply memorize it.

Please understand the argument that led to this formula, so that you will understand it more deeply. But we'll be using this formula quite a bit. Notice that one context in which evenly spaced integers arise, is the set of all positive integers that, when divided by one number, give a fixed remainder.

For example, this sequence. The sequence that we've been looking at, is the set of all positive integers that when divided by seven, have a remainder of five. That's what all those numbers have in common. When we divide by seven, we have a remainder of five. Notice that the remainder is the first term, a sub 1.

And the divisor is the common difference that will be valuable. So hold onto that thought. Here's a practice problem. Very simple practice problem. We have an arithmetic sequence.

Find the 41st term of this sequence. Pause the video and work on this. Okay. Well the first term is 14, and obviously we're adding nine each time. That's how we're advancing.

So the common difference is nine. So we can just plug into the formula. This is the general formula. If we want the 41st term, we plug in n equals 41. So we get 40, 14 plus 9 times 40.

9 times 40 of course is 360. Add 14, we get 374. 374 is the 41st term on the sequence. The 41st term on the list. Here's another practice problem.

Pause the video and then we'll talk about this. Let S be the set of all positive integers that, when divided by eight, have a remainder of five. What is the 76th number in this set? Well, remember that we said that the remainder, 5, that's gonna be the first term on the list because of course the lowest number on the, in this whole set is gonna be 5 itself.

If we divide 5 by 8, we get a quotient of 0 and a remainder of 5. And then 8 will be the common difference, because we'll be adding 8 each time to get the new numbers. So we have our starting term, we have our common difference. That's our general formula.

We plug in N equals 76 with the 76 number, we get 5 plus 8 times 75. For that 8 times 75 we'll just to doubling and halving. So it's four times 150, four times 150 is clearly 600, plus five is 605. And that is the 76th number on the list. Here's another practice problem.

This one's a little bit different. Pause the video and then we'll talk about this. Okay. Here, it turns out that the first term is not known, the common difference is not known.

We're just given two random turns somewhere on the list and we wanna find another term. Well think about it this way. Let the initial term equal b, and let the common difference equal d. Well then certainly it's true that a sub 3 would be b plus 2d and that's 17. The 19th term would be b plus 18d and that equals 65.

And what we have here are two equations with two unknowns. And in fact it's very easy to solve. We're just gonna subtract the first equation from the second equation. Subtract, we get 16D equals 48. Divide by 16, we get D equals three. And then plugging in to A3, we get that the initial term is 11.

So now we have the initial term, the common difference, we can set up the general formula. The tenth term will be initial term 11 plus common difference 3 times 10 minus 1, which is 9. 3 times 9 is 27. We add that, we get 38.

And 38 is the tenth number on the list. An arithmetic sequence is one in which the terms have a common difference. Any evenly spaced list is an arithmetic sequence. Other examples include consecutive multiples of a number, consecutive odds or evens, and numbers which when divided by the same divisor, have the same remainder.

The nth term of any arithmetic sequence, is given by the formula a sub n equals a sub 1 plus d times n minus 1.

Read full transcriptThat's how we go from one term to the next. So we could continue this pattern forever, just continually adding seven to get the next number. Notice first of all, a really big idea. Any evenly spaced list is an arithmetic sequence.

So this is powerful. One special case is consecutive multiples of an integer p. We would add p to get from each term to the next. Consecutive odd numbers, consecutive even numbers, or consecutive integers are also special cases of arithmetic sequences. Here's an arithmetic sequence.

It would be helpful to find a formula for the nth term, so that, for example, we could find the value of a much higher term in the sequence. The test likes to ask about this. Remember for the algebraically defined sequences that we saw in the previous video, it's very easy. If the test said find the 80th term in this sequence, all we'd have to do is plug in.

But we'd like a formula that would allow us to do that for arithmetic sequences. I'm gonna analyze the individual terms in the following way to look for a pattern. So I'll focus on this particular sequence. The sequence that starts with five, and then we add seven each time. So the first term is five.

The second term is 12. I'm gonna write that as five plus seven. We've added seven once. To get the third term, we add seven again. So I'm gonna write that as five plus seven plus seven. Then the fourth term, we add seven again, then add seven again.

And so what we have here, this stream of sevens. We have five plus a number of sevens. Notice that the number of sevens, in each row, the number of sevens that we're adding is one less than the index number. It's one less than the number on the list. In each term the number of seven's added as one less than the index number of the term.

This is a big idea. Thus, the nth term, a sub n, would be the starting term plus n minus one individual sevens. Of course, a compact way to add n minus 1 individual 7's would simply be to multiply 7 times n minus 1. So that means that the nth term of this sequence would be 5 plus 7 times n minus 1.

Now we can generalize a bit. First of all, the starting term in general's a sub 1. We're familiar with that from the last video. The fixed amount we add each time to get each term is called the common difference. Because if we subtract any two adjacent terms, we get this same number.

We symbolize the common difference with the letter D. The nth term of an arithmetic sequence with an initial term A sub one and a common difference D, is A sub N equals A sub one plus N minus one times D. That is a very powerful formula. Please do not simply memorize it.

Please understand the argument that led to this formula, so that you will understand it more deeply. But we'll be using this formula quite a bit. Notice that one context in which evenly spaced integers arise, is the set of all positive integers that, when divided by one number, give a fixed remainder.

For example, this sequence. The sequence that we've been looking at, is the set of all positive integers that when divided by seven, have a remainder of five. That's what all those numbers have in common. When we divide by seven, we have a remainder of five. Notice that the remainder is the first term, a sub 1.

And the divisor is the common difference that will be valuable. So hold onto that thought. Here's a practice problem. Very simple practice problem. We have an arithmetic sequence.

Find the 41st term of this sequence. Pause the video and work on this. Okay. Well the first term is 14, and obviously we're adding nine each time. That's how we're advancing.

So the common difference is nine. So we can just plug into the formula. This is the general formula. If we want the 41st term, we plug in n equals 41. So we get 40, 14 plus 9 times 40.

9 times 40 of course is 360. Add 14, we get 374. 374 is the 41st term on the sequence. The 41st term on the list. Here's another practice problem.

Pause the video and then we'll talk about this. Let S be the set of all positive integers that, when divided by eight, have a remainder of five. What is the 76th number in this set? Well, remember that we said that the remainder, 5, that's gonna be the first term on the list because of course the lowest number on the, in this whole set is gonna be 5 itself.

If we divide 5 by 8, we get a quotient of 0 and a remainder of 5. And then 8 will be the common difference, because we'll be adding 8 each time to get the new numbers. So we have our starting term, we have our common difference. That's our general formula.

We plug in N equals 76 with the 76 number, we get 5 plus 8 times 75. For that 8 times 75 we'll just to doubling and halving. So it's four times 150, four times 150 is clearly 600, plus five is 605. And that is the 76th number on the list. Here's another practice problem.

This one's a little bit different. Pause the video and then we'll talk about this. Okay. Here, it turns out that the first term is not known, the common difference is not known.

We're just given two random turns somewhere on the list and we wanna find another term. Well think about it this way. Let the initial term equal b, and let the common difference equal d. Well then certainly it's true that a sub 3 would be b plus 2d and that's 17. The 19th term would be b plus 18d and that equals 65.

And what we have here are two equations with two unknowns. And in fact it's very easy to solve. We're just gonna subtract the first equation from the second equation. Subtract, we get 16D equals 48. Divide by 16, we get D equals three. And then plugging in to A3, we get that the initial term is 11.

So now we have the initial term, the common difference, we can set up the general formula. The tenth term will be initial term 11 plus common difference 3 times 10 minus 1, which is 9. 3 times 9 is 27. We add that, we get 38.

And 38 is the tenth number on the list. An arithmetic sequence is one in which the terms have a common difference. Any evenly spaced list is an arithmetic sequence. Other examples include consecutive multiples of a number, consecutive odds or evens, and numbers which when divided by the same divisor, have the same remainder.

The nth term of any arithmetic sequence, is given by the formula a sub n equals a sub 1 plus d times n minus 1.