Lesson Video: Representing Sequences Mathematics

In this video, we will learn how to represent a sequence as a function of a positive integer variable called an index (𝑛).

17:58

Video Transcript

In this video, we’ll discover how to represent a sequence as a function of a positive integer variable called an index. We’ll look at a mixture of questions on both arithmetic and geometric sequences, including how to find the 𝑛th term of these sequence types and also how to write a sequence given the 𝑛th term. But before we begin some questions, let’s recap some of the terminology around sequences and types of sequence we might encounter.

Firstly, we have that a sequence is an ordered list of terms. Terms are usually named either π‘Ž sub 𝑖 or π‘Ž sub 𝑛, where 𝑖 or 𝑛 are the index. In this video, we’ll tend to use 𝑛 as the index. So, for example, π‘Ž sub three would be the third term. Sometimes sequences start with the index 𝑛 as 𝑛 equals one, and sometimes they start with 𝑛 equals zero. In this case, the sequence will begin with what we would call the zeroth term. And then we’d have the first term and the second term and so on. We usually get a bit of a clue in question types if they ask us to give the 𝑛th term where 𝑛 is greater than or equal to zero. In this case, we know that the index should start with 𝑛 equals zero. The terms in a sequence can be given as a list or defined by a rule often related to the index.

Let’s now recap how we would find the 𝑛th term of an arithmetic sequence and of a geometric sequence. An arithmetic sequence is a sequence with the difference between two consecutive terms constant. This difference is called the common difference. An example of an arithmetic sequence would be the sequence five, 10, 15, 20, and so on, where we can see that the common difference between any two consecutive terms would be positive five.

In order to find the 𝑛th term or π‘Ž sub 𝑛 of an arithmetic sequence, we calculate π‘Ž, the first term, plus 𝑛 minus one times d, where d is the common difference. A geometric sequence is a sequence with the ratio between two consecutive terms constant. And this ratio is called the common ratio. An example of a geometric sequence might be the sequence two, four, eight, 16, and so on. We can find the common ratio of a geometric sequence by taking any term and dividing it by the term before it. For example, if we divided the fourth term of 16 by the third term of eight, we’d get the value of two for the ratio. We’d also get this ratio of two if we divided eight by four.

To find the 𝑛th term of a geometric sequence π‘Ž sub 𝑛, we calculate π‘Ž multiplied by π‘Ÿ to the power of 𝑛 minus one, where π‘Ž is the first term and π‘Ÿ is the common ratio. Now that we’ve recapped what we know about sequences, let’s have a look at the first question, where we’ll need to establish which type of sequence we have and then find its 𝑛th term.

In any sequence pattern, if the difference between any two successive terms is a fixed number 𝑑, then this is an arithmetic sequence. Consider the sequence one, four, seven, 10, and so on and then answer the following questions. Is the sequence arithmetic? What is the value of 𝑑? What is the general term of this sequence with 𝑛 is greater than or equal to zero?

To start this question, we’re given a little reminder about the definition of an arithmetic sequence. And it’s one which has a difference or common difference between any two successive terms the same. This word successive is like the word consecutive. It simply means two terms where one immediately follows the other one. We are asked to consider the sequence one, four, seven, 10, and so on. If we look at the first question, we’re asked if the sequence is arithmetic. So if it’s arithmetic, it will have a common difference between any two consecutive terms.

So if we wanted to find the difference between the first and the second term, we would work out four take away one, which of course is three. To find the next common difference between the third term and the second term, we would work out seven take away four, and that’s also three. In the same way, the difference between 10 and seven is also three. As we have a common difference, then we do have an arithmetic sequence. And we can say yes as the answer for the first part of this question. The second part of the question asks us, what is the value of 𝑑? 𝑑 is the common difference, and we’re reminded of that in the question text. And it’s also nice and easy to work out; it’s three. And that’s the second part of the question answered.

In the final part of this question, we’re asked to find the general term of this sequence with 𝑛 greater than or equal to zero. Remember that the general term is often seen as the 𝑛th term. The fact that 𝑛 is greater than or equal to zero means that the index of this sequence should start with 𝑛 equals zero. So actually, this sequence begins with a zeroth term. Then we’d have the first term, then the second term and the third term and so on. We should recall that there is a formula to help us find the 𝑛th term of an arithmetic sequence. The 𝑛th term π‘Ž sub 𝑛 is equal to π‘Ž plus 𝑛 minus one multiplied by 𝑑, where π‘Ž is the first term and 𝑑 is the common difference.

When we look at the sequence one, four, seven, 10, and so on, the first term is really the same as π‘Ž sub one. So the value of π‘Ž, which we plug in, will be four. And as the value of 𝑑, which we worked out earlier, is three, then we add on 𝑛 minus one multiplied by three. When we distribute the three across the parentheses, we get three multiplied by 𝑛, which is three 𝑛, and three multiplied by negative one, which is negative three. Finally, when we simplify this, we get four minus three, which is one, plus three 𝑛 or three 𝑛 plus one. And that’s the answer for the third part of this question: the general term or 𝑛th term of this sequence is three 𝑛 plus one.

But it is important to note that this was because the index 𝑛 was greater than or equal to zero. If the index had started with one, i.e., 𝑛 is greater than or equal to one, then we would have worked out the 𝑛th term to be three 𝑛 minus two. So it’s really important to read the question to see if there’s an indication that the index, in fact, begins with zero. In this case, however, we can give the general term as three 𝑛 plus one.

We’ll now take a look at another question.

In a geometric sequence, the ratio between any two successive terms is a fixed ratio π‘Ÿ. Consider the sequence one-half, one-quarter, one-eighth, one sixteenth, and so on. Is this sequence geometric? Consider the sequence one-half, one-quarter, one-eighth, one sixteenth, and so on. What is the value of π‘Ÿ? Consider the sequence one-half, one-quarter, one-eighth, one sixteenth, and so on. What is the general term of this sequence?

In this question, we’re given a reminder of what a geometric sequence is. It’s one which has a fixed ratio or common ratio between any two successive terms. In each of the three parts of this question, we’re considering the same sequence. And in the first part of this question, we’re asked if this given sequence is geometric. So let’s write down this sequence. If it is geometric, then there will be a common ratio π‘Ÿ between any two consecutive or successive terms. So let’s see if we can find a ratio between the first two terms, one-half and one-quarter. To find the ratio, we take the second term of one-quarter and divide it by one-half.

When we’re dividing fractions, we write the first fraction and we multiply it by the reciprocal of the second fraction. We can take out a common factor of two from the numerator and denominator. And then multiplying the numerators gives us one, and multiplying the denominators gives us two. That means that the ratio between the first and second term is one-half. Next, we will find the ratio between the second and third term of one-quarter and one-eighth, so we’ll calculate one-eighth divided by one-quarter. Multiplying by the reciprocal of the fraction one-quarter, we calculate one-eighth multiplied by four over one. And so, once again, we get the ratio of one-half.

It looks like we probably do have a common ratio, but it’s always worth checking all the terms to make sure that there is a common ratio on all of them. And when we calculate one sixteenth divided by one-eighth, we also get the ratio of one-half. Therefore, we have established that this sequence does have a fixed or common ratio, and so it must be geometric. And we can say yes as the answer for the first part of this question. In the second part of this question, we’re asked to find the value of π‘Ÿ for the same sequence. Remember that π‘Ÿ is the fixed ratio and it’s nice and simple. We have just worked it out to be one-half, which is the second part of this question answered.

In the final part of this question, we’re asked to find the general term of this sequence. And we can remember that the general term is another way of asking for the 𝑛th term. If we started with the first term written as π‘Ž sub one equal to one-half, then the second term would be π‘Ž sub two and it would be one-quarter. The third and fourth terms can be written as π‘Ž sub three and π‘Ž sub four. So when we’re finding the general term, we’re really looking for the rule that would allow us to work out the 𝑛th term or π‘Ž sub 𝑛.

We can remember that there is a general formula to allow us to work out the 𝑛th term of any geometric sequence. π‘Ž sub 𝑛 is equal to π‘Ž multiplied by π‘Ÿ to the power of 𝑛 minus one, where π‘Ž is the first term and π‘Ÿ is the common ratio. The values that we need to plug into the formula will be π‘Ž equals one-half, as that was the first term, and π‘Ÿ we worked out as one-half. Therefore, the 𝑛th term of this sequence can be given as a half multiplied by one-half to the power of 𝑛 minus one. When we give our answer, it’s important that we indicate the values of 𝑛. When we find the 𝑛th term or general term, we started with 𝑛 equals one. So the answer for the third part of the question is that the general term of the sequence is one-half multiplied by one-half to the power of 𝑛 minus one for values of 𝑛 greater than or equal to one.

We can, of course, further simplify this general term using one of the rules of exponents. If we consider that the first value of one-half is equivalent to one-half to the power of one, and so adding the exponents one and 𝑛 minus one will give us simply the exponent of 𝑛. If we then plugged in the values of 𝑛 equals one, two, three, or four into either of these formulas, we’d get the first four terms of the sequence that we were given, and so verifying that we have the correct answer for the general term.

Next, let’s have a look at a question where we need to find the first three terms of the sequence given the general term.

Find the first three terms of the sequence whose general term π‘Ž sub 𝑛 equals 𝑛 over 𝑛 plus one.

In this question, we’re given a general term or an 𝑛th term for a sequence as 𝑛 over 𝑛 plus one. This value of 𝑛 is an index for the sequence. So when we want to work out the first term, we’d actually be working out the value of π‘Ž sub one, which means that we plug in the value of one for every value of 𝑛. Therefore, we’ll have π‘Ž sub one is equal to one over one plus one. Simplifying this gives us the value of one-half.

To find the second term or the value of π‘Ž sub two, we’ll be plugging in the value of 𝑛 equals two. This gives us two over two plus one, which when we simplify it gives us the value of two-thirds for the second term. Finally, for the third term, we’ll be working out the value of π‘Ž sub three. So we plug in 𝑛 equals three into the general term. And we have three over three plus one, which simplifies to three-quarters. Therefore, we can give the answer that the first three terms of this sequence must be one-half, two-thirds, and three-quarters.

We’ll now have a look at one final question.

Consider the sequence one, one, three-quarters, four-eighths, and so on. Which of the following is the general term of this sequence such that 𝑛 is greater than or equal to zero? Option (A) 𝑛 over two to the power of 𝑛. Option (B) 𝑛 minus one over two to the power of 𝑛. Option (C) 𝑛 plus one over two to the power of 𝑛. Option (D) two 𝑛 over two to the power of 𝑛. Or option (E) 𝑛 plus two over two to the power of 𝑛.

In this question, we’re given a sequence and asked to find its general term. When we’re finding a general term, we’re really finding a rule that connects the term number with the actual value of the term. When we’re given that the index 𝑛 is greater than or equal to zero, that means that our sequence begins with the zeroth term. We then have the first term π‘Ž sub one, the second term π‘Ž sub two, and so on. The 𝑛th term would be π‘Ž sub 𝑛. So given any value of 𝑛, what would the value in the sequence be? If we consider this sequence, there isn’t a common difference between any two consecutive terms, so this isn’t an arithmetic sequence. There also isn’t a common ratio between any two consecutive terms, so this isn’t a geometric sequence either.

In order to find the general term of the sequence, we’ll have to apply some logic. Let’s take a closer look at this term, π‘Ž sub one with the value of one. What if instead of being this value of one, this value of π‘Ž sub one was actually a fraction which simplified to one? In order for a fraction to simplify to one, the numerator and denominator would have to have the same value. Let’s say this fraction was actually two over something, and to simplify to one, it would need to be two over two. If we think of the zeroth term π‘Ž sub zero as instead of just being one as being a fraction of one over one, now we can see that the numerators actually have quite a nice pattern. They go from one to two to three to four. The denominators also have a different pattern. They go from one to two to four to eight.

Let’s consider the general term of the numerators and denominators separately for each value of 𝑛 starting with 𝑛 equals zero. Remember that we picked zero because this was given to us in the question. So for any index 𝑛, what will the numerator be? Well, every value in the numerator is one more than its index. So the 𝑛th term of the numerator will be 𝑛 plus one. For the denominators, these have a pattern which appears to be doubling. In fact, each denominator is a power of two. It would be two to the power of 𝑛. For example, for the zeroth term, two to the power of zero gives us one. For the first term, two to the power of one gives us two and so on. We can now put together the general term for the numerator and denominator. So the general term of this sequence is 𝑛 plus one over two to the power of 𝑛, which was the value given to us in option (C).

We’ll now summarize the key points of this video. Firstly, we saw that sequences consist of terms. Then we saw that terms can be written as π‘Ž sub 𝑛, where 𝑛 is the index. We also saw how sometimes the index 𝑛 begins with zero and sometimes it begins with one. We recapped arithmetic and geometric sequences and tried to find their 𝑛th terms.

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