Question Video: Using Arithmetic Mean to Solve for Unknown Values in an Arithmetic Sequence Mathematics • Second Year of Secondary School

Video Transcript

If 𝑎, 𝑏, 𝑐, 30, 𝑥, 𝑦, 𝑧 form an arithmetic sequence, then 𝑎 plus 𝑏 plus 𝑐 plus 𝑥 plus 𝑦 plus 𝑧 equals blank.

In an arithmetic sequence, consecutive terms are separated by a common difference. We often represent that with the letter 𝑑. And here we could show that to get from term one to term two, we add a common difference of 𝑑. 𝑎 plus 𝑑 equals 𝑏. And this pattern continues. 𝑏 plus 𝑑 equals 𝑐. And 𝑐 plus 𝑑 equals 30. This pattern continues throughout the sequence.

Let’s consider two different ways we could solve this problem. Our ultimate goal is to be able to add six of the values from the sequence together. One way to do that would be to write all of the variables in terms of 30 and 𝑑. Moving to the left of 30, we can rewrite 𝑐 as 30 minus 𝑑. Because we know that 𝑐 plus 𝑑 equals 30, 30 minus 𝑑 will equal 𝑐. We can move to the left a further time. And of course, we could say 𝑐 minus 𝑑 equals 𝑏. But we can actually say 𝑏 is equal to 30 minus two 𝑑. If we start at 30 and go to 𝑏, we need to subtract two 𝑑, which means that 𝑎 will be equal to 30 minus three 𝑑.

Doing this with 𝑥, 𝑦, and 𝑧, 𝑥 equals 30 plus 𝑑, 𝑦 equals 30 plus two 𝑑, and 𝑧 equals 30 plus three 𝑑. And then we’ll substitute these values in for our six variables, which will look like this. Since we’re only dealing with addition and subtraction, we don’t need those parentheses. To combine like terms, we see that negative three 𝑑 plus three 𝑑 equals zero, negative two 𝑑 plus two 𝑑 equals zero, and negative 𝑑 plus 𝑑 equals zero, leaving us with 30 plus 30 plus 30 plus 30 plus 30 plus 30, or 30 times six, which is 180.

But let’s go back and consider something else. Based on the properties of arithmetic sequences, we can know for sure that 30 is halfway between 𝑐 and 𝑥. That is, the distance from 𝑐 to 30 will be equal to the distance from 30 to 𝑥 on a number line. And this makes 30 the arithmetic mean between 𝑐 and 𝑥. And if 30 is the arithmetic mean of 𝑐 and 𝑥, then we can set up 𝑐 plus 𝑥 divided by two equals 30. And then when we multiply both sides of this equation by two, 𝑐 plus 𝑥 has to equal 60.

And then we can notice something interesting. The distance from 𝑏 to 30 equals two 𝑑 and the distance from 30 to 𝑦 is two 𝑑, which means 30 is also the arithmetic mean of 𝑏 and 𝑦, which makes 𝑏 plus 𝑦 divided by two equal to 30. And we can rewrite that as 𝑏 plus 𝑦 equals 60. And finally, because the distance from 𝑎 to 30 equals three 𝑑 and the distance from 30 to 𝑧 equals three 𝑑, 30 is the arithmetic mean of 𝑎 and 𝑧, which means 𝑎 plus 𝑧 has to equal 60.

Combining these three equations, on the left, would give us 𝑎 plus 𝑏 plus 𝑐 plus 𝑥 plus 𝑦 plus 𝑧 and, on the right, would give us 180. The six variables in this arithmetic sequence would combine to equal 180. Now we do have enough information to find what each individual variable would be. We would have to do a bit more with these equations. However, as this question has only asked the sum of these six variables, we can stop here.