Lesson Video: Calculations with Arithmetic Sequences Mathematics • Second Year of Secondary School

Video Transcript

In this video, we’re going to look at arithmetic sequences. First, we’ll look at the definition and then consider some typical arithmetic sequences. We’ll be analyzing the differences between consecutive terms, looking for the value of the zeroth term in order to find an 𝑛th term formula. The 𝑛th term formula will allow us to find any term in the sequence.

So let’s look at the numbers five, eight, 11, 14, 17. We know this is a sequence as it is an ordered list of numbers. A term of a sequence is each number in a sequence. And we number the terms consecutively. The first term of this sequence is five. The second term is eight. The third term is 11. And it could continue in this pattern. This numbering continues indefinitely as the sequence can continue indefinitely. The term number tells us the position of that value in the sequence as well. This means we could say 11 is the third term in the sequence.

As we move from one term to the next, we’re adding three to the previous term to get the next consecutive term. We’re adding three every time. This property, adding three each time, means that this sequence is an arithmetic sequence. In an arithmetic sequence, the difference between any two consecutive terms is always the same. We can label our sequence as such. The difference between any two consecutive terms in an arithmetic sequence is called the common difference. In our example, the common difference is positive three.

We should note here that the first term in an arithmetic sequence is important. If we know the first term and the common difference — in our case, positive three — we can work out any term in the sequence. The first term and the common difference are two key bits of information when dealing with any arithmetic sequence. To sum this up, if you spot a sequence where consecutive terms have a common difference, you can say it’s an arithmetic sequence.

Once we know what an arithmetic sequence is, we can solve questions like this one.

Write the next three terms in the sequence: 31, 57, 83, 109, continuing.

We’ve been given the first, second, third, and fourth terms in a sequence, which means we need to find the fifth, sixth, and seventh term. So we ask ourselves, how do we get from 31 to 57? We add 26. And 57 plus 26 equals 83. 83 plus 26 equals 109. This makes positive 26 the common difference. And it means to get from the fourth to fifth term, we’ll need to add 26 to 109. 109 plus 26 equals 135. 135 plus 26 equals 161. And to find the seventh term, we need to add 26 to the sixth term. 161 plus 26 equals 187. And so we can say that the next three terms in this sequence are 135, 161, and 187.

Let’s try another one.

Write the next three terms in the sequence: 3.6, 4.3, 5.0, 5.7.

We’ve been given the first four terms. The next three terms would be the fifth, sixth, and seventh term. To go from 3.6 to 4.3, it’s an increase of 0.7, an increase of seven-tenths. From 4.3 to five is an increase of 0.7, and from five to 5.7 is an increase of 0.7, which means we have a common difference of positive 0.7 and confirms that this is an arithmetic sequence. So we know to find the fifth term, we need to add the common difference to the fourth term. 5.7 plus 0.7 equals 6.4, 6.4 plus 0.7 equals 7.1, and 7.1 plus 0.7 equals 7.8. The next three terms in this sequence would be 6.4, 7.1 and 7.8.

What we’re seeing now is that arithmetic sequences can occur in all kinds of types of values, in integers and decimals but also in fractions. As long as the common difference between consecutive terms is the same, it’s an arithmetic sequence. Let’s look at another example.

Write the 10th term in the arithmetic sequence 23, 19, 15, 11, 7.

We’ve been given terms one through five. We’re not worried about terms six through nine. We only need to find the 10th term. The first thing we should notice is that the sequence is decreasing. Each term is four less than the previous term. And so the common difference is negative four. Let’s consider two different ways to work out the 10th term.

We could solve for the sixth, seventh, eighth, and ninth term and then find the 10th term. We would need to subtract four from each of the previous terms. Or we could start at the fifth term, seven, and then realize that we need to add five lots of negative four to get to the 10th term. In the first way, seven minus four is three, three minus four is negative one, negative one minus four is negative five, negative five minus four is negative nine, and negative nine minus four is negative 13.

Using the other method, we would start at seven, and we would add five lots of negative four. So seven plus five times negative four equals seven minus 20. Five times negative four is negative 20. And seven minus 20 is negative 13. Both methods show the 10th term is negative 13.

But what if our question had asked us to write the 250th term of the sequence? We wouldn’t want to work out all 250 of the terms. So let’s consider a way to find a general formula that will give us any term in the sequence.

If we have a sequence three, seven, 11, 15, the common difference is positive four. We’re looking for the general form. We can let the letter 𝑛 stand for the term number. That is the position in our sequence. We can use the notation 𝑡 and parentheses 𝑛 to represent the 𝑛th term. When 𝑛 is one, 𝑡 one is three. When 𝑛 is two, 𝑡 of two equals seven, and so on. If we look closely, we see every time our position in the sequence goes up by one, the value of the term in the sequence goes up by four. The value of the term is going up four times faster than the value of the position. And so we know that part of our formula will have plus four 𝑛.

If positive four 𝑛 was our general formula, we would say 𝑡 of 𝑛 equals four 𝑛. When 𝑛 equals one in the first position, 𝑡 of 𝑛 would equal four. When 𝑛 equals two, 𝑡 of 𝑛 would be equal to eight. When 𝑛 equals three, 𝑡 of 𝑛 would be equal to 12. And when 𝑛 equals four, 𝑡 of 𝑛 would be equal to 16. But this formula is not exactly right. When 𝑛 equals one, we should have three and not four. When 𝑛 equals two, we should have seven and not eight. When 𝑛 equals three, we should have 11 and not 12. And when 𝑛 is four, we should have 15 and not 16. All of these values were one off. Four minus one is three, eight minus one is seven, 12 minus one is 11, and 16 minus one is 15. And so we can add minus one to our 𝑡 of 𝑛 formula so that the new general formula is 𝑡 of 𝑛 is equal to positive four 𝑛 minus one, but we don’t have to include the positive sign.

The formula for finding the 𝑛th term of this sequence will be 𝑡 of 𝑛 equals four 𝑛 minus one. If we use this form to calculate the 250th term, four times 250 is 1000, 1000 minus one is 999. The 250th term in this sequence is 999. We should also point out that there are a few different notations for 𝑡 of 𝑛. You might use 𝑡 sub 𝑛 or even 𝑎 sub 𝑛. All three of our ways to represent the 𝑛th term.

Let’s try that method on a new problem.

Find the 81st term in the sequence 107, 99, 91, 83.

We’ve been given the first four terms in a sequence. We’ll let 𝑛 be the position number or the term number and 𝑡 of 𝑛 is the value of the term in the sequence. We recognize that the sequence is going down by eight; our common difference is negative eight. To make a general formula for this sequence, we would need negative eight 𝑛. If we plug in our 𝑛 values one through four, we would end up with the values negative eight, negative 16, negative 24, negative 32. But when 𝑛 equals one, our value is 107 not negative eight. When 𝑛 equals two, we need 99 not negative 16. For the third term, we need 91 and not negative 24. And for the fourth term, we need 83 not negative 32.

If we add 115 to negative eight, we would get 107. If we add 115 to negative 16, we would get 99. And so we need to add 115 to the negative eight 𝑛. A general formula for this sequence would look like this. 𝑡 of 𝑛 is equal to negative eight 𝑛 plus 115. To solve for the 81st term in the sequence, we would need 𝑡 of 81, which is equal to negative eight times 81 plus 115. And that’s negative 533. The 81st term in this sequence is negative 533.

Finding the general form with this method still requires a lot of calculation, so let’s consider another way to do that. If we take our 𝑛 and 𝑡 of 𝑛 values and plot them along an 𝑥- and 𝑦-axis, we’ll start with the table that looks something like this. Plotting those points would give us something that looks like this. 𝑛 can only take integer values, one, two, three, four, five, and so on. Because our sequence has terms one, two, three, four, and so on. It doesn’t make sense to talk about the 3.75 term, but let’s consider what would happen if we connected these points with a line.

The line would cross the 𝑦-axis at 115. This is because when we increase our 𝑥-value by one, the 𝑦-value decreases by eight. And this means if we do the opposite — if we decrease our 𝑥-value by one — we need to increase our 𝑦-value by eight. If we move to the left from 𝑥 equals one, we get the place where 𝑥 equals zero. And we need to increase by eight. And we see that when 𝑥 equals zero, 𝑦 equals 115. If we were talking about this like a line, we would say it has a slope of negative eight and a 𝑦-intercept of 115. We would say that 𝑦 equals negative eight 𝑥 plus 115.

And we can use this to help us find the general form in an arithmetic sequence. The common difference is negative eight. The common difference told us the slope of the line. And the 𝑦-intercept told us the number we needed to add. This matches the form we found previously, 𝑡 of 𝑛 equals negative eight 𝑛 plus 115. Let’s try and use this method on a different problem.

Find the formula for the 𝑛th term of the arithmetic sequence 6.8, 7.9, 9.0, 10.1.

We’re given the first four terms in an arithmetic sequence. We’ll let the term number be equal to 𝑛 and 𝑡 of 𝑛 is the corresponding value of the term in the sequence. To get from term one to term two, we add 1.1. To get from term two to three, we add 1.1. And term three plus 1.1 equals term four. The common difference is positive 1.1. What we want to do now is find the value of our zero term.

When we move along consecutively from left to right, we’re adding 1.1. If we want to move in the opposite direction, we would be subtracting 1.1. To find our zeroth term, we need to subtract 1.1 from our first term. 6.8 minus 1.1 equals 5.7. The formula for the 𝑛th term of this sequence equals 1.1 times 𝑛 plus 5.7.

Let’s summarize what we’ve just shown. We could say that the formula for the 𝑛th term of an arithmetic sequence looks like this. 𝑡 of 𝑛 equals 𝑎 times 𝑛 plus 𝑏. The constant 𝑎 is the value we multiply by the term number, the common difference, which can be positive or negative. And the 𝑏 value is the constant that is the zeroth term of the sequence. If we see formulas with squares, square roots, or powers of 𝑛, the sequence is not an arithmetic sequence. All of these are nonlinear equations and, therefore, cannot represent an arithmetic sequence.

Let’s consider a final type of question you might see when dealing with arithmetic sequences.

Does 117 belong to the arithmetic sequence five, 18, 31, 44?

We’re given the first four terms in an arithmetic sequence. We’ll let 𝑛 be the term number and 𝑡 of 𝑛 is the value of the term in the sequence. The first thing we should check here is the common difference. To get from term to term here, we’re adding 13. And the common difference is then positive 13. At this point, it’s good to remember 𝑡 of 𝑛 equals 𝑎 times 𝑛 plus 𝑏, where 𝑎 is the common difference and 𝑏 is the zeroth term. And this means the next thing we need to find is what the zero term is.

In this sequence, when we’re moving to the right, we’re adding 13. And that means to move to the left, we need to subtract 13. Five minus 13 is negative eight. And negative eight is the zero term. We now have enough information to write the general form for the 𝑛th term of this sequence. That will be 13𝑛 minus eight, a common difference of positive 13 and a zeroth term of negative eight.

In order to find out if 117 belongs to this sequence, we plug 117 in for 𝑡 of 𝑛. If 117 is part of this sequence, it will have an integer value for 𝑛. So we solve this equation for 𝑛 and see if it’s an integer. To do that, we first add eight to both sides. This gives us 125 is equal to 13𝑛. And so we divide both sides of the equation by 13. 125 divided by 13 is equal to 9.6153 continuing or, as a fraction, nine and eight thirteenths. Because 9.6153 continuing is not an integer, 117 is not found in this sequence. When dealing with arithmetic sequences and this general form, the 𝑛-value must be an integer.

To summarize what we know, an arithmetic sequence is a sequence in which the difference between any two consecutive terms is always the same. The common difference is the difference between any two consecutive terms in an arithmetic sequence. And we can use the general form 𝑡 of 𝑛 equals 𝑎𝑛 plus 𝑏, where 𝑎 equals the positive or negative common difference and 𝑏 equals the zeroth term.