Okay, let me start by saying that I’m not sure how this would be taught in a classroom, since it’s not something I ever taught myself, but this is how I figured it out myself.
In the case of your first sequence, I’ll assume you left out the last comma (the one before the 10), and that it should read 2, __, __, __, 10. This means that there are three missing terms in the sequence. Since that’s an odd number of missing terms, I realized that the middle term must be the arithmetic mean of the first and last terms, 2 and 10. (2 + 10)/2 = 12/2 = 6, which has to be the middle term. (It will also be our final answer for the sequence. I would have stopped here, but let’s demonstrate how that works.)
The sequence now reads 2, __, 6, __, 10. Let’s fill in the blank between the 2 and the 6 by finding their arithmetic mean, just as we did above.
(2 + 6)/2 = 8/2 = 4. So now we know the sequence is 2, 4, 6, __, 10.
Using the same reasoning, we can fill in the final blank by finding the arithmetic mean between 6 and 10. (You can probably guess this answer, but let’s do the math anyway, since future problems might not be that easy to fill in.)
(6 + 10)/2 = 16/2 = 8. The sequence is complete: 2, 4, 6, 8, 10.
The arithmetic mean of this sequence is (2 + 4 + 6 + 8 + 10)/5 = 30/5 = 6.
Notice that, when there are an odd number of terms in the sequence, the arithmetic mean of the sequence is simply the arithmetic mean of the first and last terms. But is that true when there is an even number of terms, as there is in the second sequence?
Again, this is how I think about it, not necessarily how I would teach it.
In the sequence 13, __, __, 40, we have two missing terms. simply finding the mean of 13 and 40 would give us (13 + 40)/2 = 53/2 = 26.5, but is that the arithmetic mean of the sequence?
I decided to figure out what the missing terms were to check.
I examined the sequence 13, __, __, 40. To explain what I did next, I’m going to use some variables. Think of the sequence as 13, x, y, 40. If you aren’t using variables yet, don’t worry. Writing letters there just makes it easier for me to explain what I did in my head. We won’t be using algebra here.
I examined the sequence and noticed that there are three “jumps” from the first term to the last. What I mean is that there is one jump from the 13 to the x, another jump from the x to the y, and a third jump from the y to the 40.
Since each of those jumps must be the same (that’s what sequences are all about, right?), then it takes three jumps to get from 13 to 40.
But how big is the gap between 13 and 40? Let’s subtract: 40–13 = 27, so the gap between 13 and 40 is 27. Since we know that there are 3 “jumps” to take us from 13 to 40, let’s divide 27 by three to see how big each of those jumps is. 27/3 = 9. Each jump is 9, so let’s fill in the sequence.
13 + 9 = 22, so the first blank (where I put the x) is 22.
Then 22 + 9 = 31, so the second blank (where I put the y) is 31.
By the way, notice that if we do the final jump, 31 + 9 = 40, which is exactly what we’d hoped!
Finally, since our sequence now reads 13, 22, 31, 40, we can find the arithmetic mean of all four numbers in the sequence:
(13 + 22 + 31 + 40)/4 = 106/4 = 26.5, which is exactly the same answer as what happened when we just added 13 and 40 and divided by 2.
Well, that was a lot of work, but it saves us some time in the future! Now we know that if we have an arithmetic sequence, all we have to do to find the arithmetic mean of that sequence is to add the first and last terms and divide by 2. (We don’t have to worry about all those “jumps” that I was talking about earlier.)
I hope this helps.
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