Lesson Video: Arithmetic Mean Mathematics

Video Transcript

In this video, we will learn how to find the arithmetic means for any two nonconsecutive terms in an arithmetic sequence.

Let’s start by defining arithmetic mean. An arithmetic mean is the sum of a set of values divided by the number of values in the set. If we have two numbers three and nine and we want to find their arithmetic mean, we add them together, three plus nine, and then divide by two because we have a set of two numbers. And this equals six. What we’re showing here is that the distance from three to six is equal to the distance from six to nine. Let’s call that distance 𝑑. We can see that from three to six we’re adding three and from six to nine we’re adding three.

If we continue this pattern, nine plus three is 12, and 12 plus three is 15. We could call this set of values an arithmetic sequence. This is because in an arithmetic sequence, the difference between any two consecutive values will be equal. But we wanna primarily focus in this video on the arithmetic means. And in this sequence, there is more than one arithmetic mean. We’ve shown how six is the arithmetic mean between three and nine. But nine is also an arithmetic mean in this sequence. If we take the values on either side, we can say six plus 12 divided by two equals nine. And that makes nine the second mean in this sequence, which would make 12 the third mean. And that means we can say that there are three arithmetic means between three and 15.

Using our knowledge about the behavior of arithmetic sequences and what we know about arithmetic means inside arithmetic sequences, let’s look at some example questions.

Find five arithmetic means between seven and 19.

We have two values, seven and 19. And we’ve been told that there are five arithmetic means between them, which we could mark 𝑎, 𝑏, 𝑐, 𝑑, 𝑒. Our first mean is 𝑎. And we know that if 𝑎 is the first mean, the distance from seven to 𝑎 must be equal to the distance from 𝑎 to 𝑏. And so we can say if seven plus 𝑥 equals 𝑎, then 𝑎 plus 𝑥 must equal 𝑏. Moving on, 𝑏 is the second mean, which means that the distance from 𝑎 to 𝑏 must be equal to the distance from 𝑏 to 𝑐. If 𝑎 plus 𝑥 equals 𝑏, then 𝑏 plus 𝑥 must equal 𝑐. This must be true for all five of our arithmetic means between seven and 19. They must have a common difference. And we’re calling that common difference 𝑥.

If there’s a common difference of 𝑥 between all five of these arithmetic means, there will also be a common difference of 𝑥 between the last mean 𝑒 and the end of our sequence 19. We can use this information to set up an equation. We can create a relationship between seven and 19 using the common differences. To get from seven to 19, if there are five arithmetic means between them, you need to add six 𝑥. So we’re saying seven plus six 𝑥 must be equal to 19. This equation will allow us to solve for that common difference.

We subtract seven from both sides of our equation and find six 𝑥 equals 12, which means 𝑥 equals two. And if 𝑥 equals two, 𝑎 equals nine, 𝑏 equals 11, 𝑐 equals 13, 𝑑 equals 15, and 𝑒 equals 17. And the five arithmetic means between seven and 19 are nine, 11, 13, 15, and 17.

In our next example, we’ll look at what to do if we’re given the sum of different means in an arithmetic sequence.

If the sum of the second mean and the fourth mean from an arithmetic sequence equals 16 and the seventh mean is more than the third mean by eight, then the sequence is blank.

Let’s say we have some sequence with the first term 𝑎. We could say the second term is 𝑏, the third term is 𝑐, and continue on like this. We have to remember that if the first term is 𝑎, the first mean is actually the second term in the sequence. If the second term is the first mean, the third term is the second mean and the fifth term is the fourth mean. However, this string of variables is not very helpful to us. It would be more helpful to write these variables in terms of our first value in the sequence.

So let’s go back. If we let our first value in the sequence be 𝑎, we know that our second term will be equal to our first term plus a common difference. And this is a much better way to write these values. Our second term would be equal to 𝑎 plus the common difference 𝑑. And our third term would be equal to 𝑎 plus 𝑑 plus 𝑑. We can call that 𝑎 plus two 𝑑. Our fourth term would be equal to 𝑎 plus three 𝑑. And we need to think about the values we are interested in. We have information about the second mean, the fourth mean, the seventh mean, and the third mean.

We’ve already said that that second mean would be equal to the third term and the fourth mean would be equal to the fifth term. We can notice something interesting here. The second mean has two units of the common difference 𝑑 being added to the first term. And the fourth mean has four common differences being added to the first term 𝑎. We see that with the third mean. Three common differences are being added. And we can use that to say that the seventh mean would be the first term plus seven 𝑑.

Using these values, we can set up some simultaneous equations to solve for the sequence. We know the second mean plus the fourth mean equals 16. This means 𝑎 plus two 𝑑 plus 𝑎 plus four 𝑑 equals 16. We also know that the third mean plus eight is equal to the seventh mean. So we write 𝑎 plus three 𝑑, that’s the third mean, plus eight is equal to 𝑎 plus seven 𝑑, the seventh mean. On the left, we can simplify to two 𝑎 plus six 𝑑 equals 16. And for our other equation, if we subtract 𝑎 from both sides, we have 𝑎 minus 𝑎 on both sides.

We end up with three 𝑑 plus eight equals seven 𝑑. By subtracting three 𝑑 from both sides, we can see that eight is equal to four 𝑑. And by dividing both sides by four, we find out that two equals 𝑑 or 𝑑 equals two. And then we wanna take that two for our 𝑑-value and plug it into our second equation so that we have two 𝑎 plus six times two equals 16. Two 𝑎 plus 12 equals 16. By subtracting 12 from both sides, we get two 𝑎 is equal to four. And once we divide through by two, we see that 𝑎 equals two.

Remember that 𝑎 represented the first term in our sequence and our common difference here, our 𝑑-value, is two. This means that our second term will be four, our third term will be six, and the pattern would continue. The arithmetic sequence described here is the sequence with the first term of two and a common difference of two.

Our next example is a little different. We want to find the number of means between two given values if we’re given information about some of the arithmetic means in between the two values.

Find the number of arithmetic means inserted between eight and 238 given the sum of the second and the sixth means is 96.

Let’s think about what we know. We’re given eight and 238. And we’re trying to figure out how many arithmetic means are between these two values given the sum of the second and the sixth means is 96. We don’t know anything about the terms between eight and 238 apart from that, but we do know that every consecutive term will have a common difference. Our second mean is two common differences away from eight. The second mean is the third term. So let’s let 𝑎 be equal to our second mean.

If 𝑎 equals our second mean, it’s going to be equal to eight plus two times the common difference 𝑑. To go from our second mean to our sixth mean, we need to add that common difference four more times. So we’ll let the sixth mean be equal to 𝑏. We can write 𝑏 in terms of our first term and our common difference. 𝑏 would be equal to eight plus six 𝑑. If you start at eight and try to get to 𝑏, you need to add the common difference six times. We know that the sum of the second and sixth means is 96. 𝑎 plus 𝑏 must be equal to 96. We can plug in eight plus two 𝑑 for 𝑎 and eight plus six 𝑑 for 𝑏.

When we combine like terms, we find that 16 plus eight 𝑑 equals 96. Subtracting 16 from both sides gives us eight 𝑑 is equal to 80. And dividing both sides by eight, we get that 𝑑 equals 10. This is not telling us how many arithmetic means are between eight and 238. It’s only saying that the common difference in this sequence is 10.

Now, we need to think of a way to go from eight to 238 with a common difference of 10. We want to know if we start with eight, how many sets of 10 do we need to add to eight to end up at 238? We subtract eight from both sides, and we get 𝑥 times 10 equals 230. If we divide both sides by 10, we get that 𝑥 is equal to 23. This means we’re saying eight plus 23 times the common difference equals 238. And that makes sense. The common difference is 10, so eight plus 230 equals 238. But here’s where we have to be really careful. This 23𝑑 gets us from the first term to the last term, but our question wants to know the number of arithmetic means between 238 and eight. And this means we need to go one to the left of 238.

To get from 238 to the final mean, we subtract 𝑑. If we let 𝑐 be equal to the final mean between eight and 238, it’s located at eight plus 23𝑑 minus 𝑑. It would be equal to eight plus 22𝑑. And that 22 makes it the 22nd mean, which means there are 22 means inserted between eight and 238.

In our final example, we’ll again be trying to find the number of arithmetic means inserted between two values. But this time, we’re given the ratio between two different sets of means.

Find the number of arithmetic means inserted between two and 254 given the ratio between the sum of the first two means and the sum of the last two means is 11 over 245.

Let’s think about what we know. We have some sequence where the first term is two and the last term is 254. In sequences like this, the second term is equal to the first mean and the third term is equal to the second mean. We’ll let 𝑎 be our first mean and 𝑏 be our second mean. We know to get from our first term to our second term, there must be a common difference of 𝑑. The same thing is true. To get from our second term to our third term, we need to add a common difference of 𝑑. But how should we label our last two means?

If we began at our final term 254, the last mean will be negative 𝑑 away from the last term. We can let 𝑒 be our last mean. And if we take the last mean and subtract the common difference of 𝑑, we get a second to last mean. We’ll write our first two means in relation to the first term. 𝑎 will be equal to two plus 𝑑. And 𝑏 would be equal to two plus two 𝑑. And then we can write 𝑒 and 𝑓 in terms of our last value 254. 𝑒 will be equal to 254 minus 𝑑. And 𝑓, the second to last term, would be equal to 254 minus two 𝑑.

Now our ratio is the sum of the first two over the sum of the second two. And that means we want to add 𝑎 and 𝑏 and add 𝑒 and 𝑓. For the first two means, they sum to four plus three 𝑑. And for the last two means, they sum to 508 minus three 𝑑. We’ll take these two expressions and set them equal to our ratio of 11 over 245. We have the sum of the first two means, four plus three 𝑑, over the sum of the last two means, 508 minus three 𝑑, which must be equal to 11 over 245.

When we cross multiply, we get 245 times four plus three 𝑑 is equal to 11 times 508 minus three 𝑑. So we distribute, which gives us 980 plus 735𝑑 on the left and 5588 minus 33𝑑 on the right. Next, we add 33𝑑 to both sides of the equation. And then we need to subtract 980 from both sides to get 768𝑑 is equal to 4608. When we divide both sides of the equation by 768, we find that 𝑑 equals six. We now know the common difference is six. And we’ll need to use this to figure out how many means are between two and 254. That means we’ll need to figure out how do we get from two to 254 in increments of six.

Algebraically, we can write that as two plus 𝑥 times six is equal to 254 and then solve for 𝑥. When we do that, we find that 𝑥 equals 42. That means that we are taking 42𝑑 and adding it to our first term of two to get to 254. But to get to the last mean, we only need to add 41𝑑. Because two plus 41𝑑 equals the last mean, there are 41 means between two and 254.

Before we finish, let’s review our key points. The terms between two nonconsecutive terms in an arithmetic sequence are known as arithmetic means. To calculate arithmetic mean, take the sum of a group of values and divide by the number of values. Using these properties, you can solve for the number of arithmetic means between two arbitrary values.