Lesson Video: Geometric Sequences | Nagwa Lesson Video: Geometric Sequences | Nagwa

Lesson Video: Geometric Sequences Mathematics

In this video, we will learn how to calculate the common ratio, find next terms in a geometric sequence, and check if the sequence increases or decreases.

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Video Transcript

In this video on geometric sequences, we will learn how to calculate the common ratio between terms, find the next terms in a geometric sequence, and check if the sequence increases or decreases.

But let’s start by thinking about what a geometric sequence actually is. One of the geometric sequences we might commonly see is the sequence one, two, four, eight, 16, and so on. Each term in the sequence is double the term before it. We can say that the ratio between any two successive terms in the sequence is two. We can see a whole range of different geometric sequences. For example, we could have sequences with noninteger values and sequences which have a negative ratio between successive terms. These type of sequences are referred to as alternating geometric sequences because the signs switch between positive and negative.

But what defines a geometric sequence is that the ratio is common between successive terms. We can use the terminology that the sequence can be given as the terms π‘Ž sub one, π‘Ž sub two, π‘Ž sub three, π‘Ž sub four, and so on. We may also see this with alternative letters other than using π‘Ž, for example, the sequence defined as 𝑑 sub one, 𝑑 sub two, and so on. Some sequences may begin with π‘Ž sub zero, but we’re really just giving a position value to each term.

We can define a geometric sequence then as a sequence of nonzero numbers π‘Ž sub one, π‘Ž sub two, π‘Ž sub three, π‘Ž sub four, and so on that has a nonzero common ratio π‘Ÿ, which is not equal to one, between any two consecutive terms. The common ratio π‘Ÿ is equal to π‘Ž sub 𝑛 plus one over π‘Ž sub 𝑛 for values of 𝑛 equal to one, two, three, and so on. This part of the equation π‘Ž sub 𝑛 plus one over π‘Ž sub 𝑛 simply indicates any term in the sequence divided by the term immediately before it.

If we take this example sequence below, we can say that the ratio π‘Ÿ is equal to π‘Ž sub two over π‘Ž sub one. That’s 100 divided by negative 10, which is negative 10. But we could have equally found the ratio by dividing π‘Ž sub four by π‘Ž sub three. And that would also give us the same ratio of negative 10. But before we look at some questions, we can also define increasing and decreasing geometric sequences.

We say that a geometric sequence is increasing if π‘Ž sub 𝑛 plus one is greater than π‘Ž sub 𝑛, and it would be decreasing if π‘Ž sub 𝑛 plus one is less than π‘Ž sub 𝑛. So it’s increasing if each term is greater than the term before it and decreasing if a term is less than the term before it. In each case, this has to be true for all index values of 𝑛.

We can now take a look at some examples on geometric sequences, and we’ll begin with one in which we need to find the common ratio.

The table shows the number of bacteria in a laboratory experiment across four consecutive days. The number of bacteria can be described by a geometric sequence. Find the common ratio of this sequence.

In the table, we can see that on four different days, there is a number of bacteria recorded in this experiment. We are told that the number of bacteria forms a geometric sequence and that is one which has a common ratio between any two consecutive terms. In order to find this ratio, then, we could take any term and divide it by the term before it. For example, we could take the second term, which could be denoted as π‘Ž sub two, and divide it by the first term.

This would be 2572 divided by 643. Putting this into our calculators or simplifying the fraction, we would get an answer of four.

As a check of our answer, we could find the ratio between another pair of consecutive terms. For example, we could take the fourth term and divide it by the third term. Simplifying 41152 over 10288 would also give us an answer of four. Because we know that this is a geometric sequence, we don’t need to do both sets of calculations. The second one was a good check, however. We can give the answer, then, that the common ratio of this sequence is four.

In the next example, we’ll see how we can find the next term of a geometric sequence.

Find the next term of the geometric sequence negative five, negative five over four, negative five over 16, negative five over 64, what.

Because we are given that this sequence is geometric, that means we know that there is a common ratio between consecutive terms. In other words, there is some ratio π‘Ÿ which we can multiply any term by to get the next term. And so, in order to find this value of π‘Ÿ, we can take any term in the sequence, which we can call π‘Ž sub 𝑛 plus one, and divide it by the term before, which we can denote as π‘Ž sub 𝑛.

We can take two terms in the sequence which might be the easiest to divide. So let’s take the second term and divide it by the first term. So we have negative five over four over negative five. And it’s helpful to remember that this is the same as negative five over four divided by negative five. If we take negative five as the fraction negative five over one, then we can remember that to divide by a fraction, we multiply by its reciprocal.

We can take out a common factor of five and then observe that we have a negative fraction multiplied by a negative fraction. And so we’ve worked out that the ratio π‘Ÿ is one-quarter.

This means that to work out the fifth term, we multiply the fourth term by the fraction of one-quarter. So we work out negative five over 64 multiplied by one-quarter. This gives us negative five over 256. We can’t simplify this fraction any further, so we can give the answer that the next term of this geometric sequence is negative five over 256.

In the next example, we will find the common ratio of a geometric sequence which is defined in a recursive formula.

Find the common ratio of the geometric sequence which satisfies the relation π‘Ž sub 𝑛 equals nine-eighths π‘Ž sub 𝑛 plus one, where 𝑛 is greater than or equal to one.

Let’s begin by recalling that a geometric sequence is one which has a common ratio π‘Ÿ between any two consecutive terms. We can find this common ratio by dividing any term, which we denote as π‘Ž sub 𝑛 plus one, by the term immediately before it, which we denote as π‘Ž sub 𝑛. This is usually easy to do if we’re actually given the terms in the sequence, but we aren’t in this question. But we are given a relationship between π‘Ž sub 𝑛 and π‘Ž sub 𝑛 plus one.

So let’s look at this statement, this equation that we have written regarding the common ratio. If we multiply both sides of this equation by π‘Ž sub 𝑛, we get the equation π‘Ÿ times π‘Ž sub 𝑛 equals π‘Ž sub 𝑛 plus one. By dividing through by π‘Ÿ, we get that π‘Ž sub 𝑛 equals π‘Ž sub 𝑛 plus one over π‘Ÿ. We can then relate this to the relation in the question, which is written in terms of π‘Ž sub 𝑛. We can set the right-hand side of both of these equations equal to each other, so we have π‘Ž sub 𝑛 plus one over π‘Ÿ equals nine-eighths π‘Ž sub 𝑛 plus one.

Dividing both sides of this equation by π‘Ž sub 𝑛 plus one, we have one over π‘Ÿ equals nine-eighths. And so the answer is that the common ratio π‘Ÿ is eight-ninths.

This question can be a little tricky to understand, especially if we’re wondering why the ratio isn’t just nine over eight. So let’s consider this question as a diagram. Imagine that we have this sequence, and we don’t know the values in the sequence. We do, however, know a relationship between a term π‘Ž sub 𝑛 and π‘Ž sub 𝑛 plus one. But we’re almost given the relationship in the wrong direction; we’re told how to get π‘Ž sub 𝑛 from π‘Ž sub 𝑛 plus one. We multiply π‘Ž sub 𝑛 plus one by nine over eight to get π‘Ž sub 𝑛.

But when we’re thinking about sequences, we think about how we go from one term to the term after that term. The inverse of multiplying by nine over eight is dividing by nine over eight. But when we’re giving a ratio, it must be in terms of a multiplier. That’s the reciprocal. So here we would multiply by eight-ninths. And that’s why the common ratio of this sequence is eight over nine.

In the next example, we’ll see how we can generate the first terms of a sequence given its general term.

Find the first five terms of the sequence π‘Ž sub 𝑛 given π‘Ž sub 𝑛 plus one equals one-quarter π‘Ž sub 𝑛, 𝑛 is greater than or equal to one, and π‘Ž sub one equals negative 27.

In this question, we’re given the information which we can use to generate the terms of this sequence. Although this notation of π‘Ž sub 𝑛 plus one and π‘Ž sub 𝑛 can look confusing, all that this formula is telling us is that if we wish to generate any term in the sequence, then we take the term before it and multiply it by one-quarter. The first five terms in the sequence can be given as π‘Ž sub one, π‘Ž sub two, π‘Ž sub three, π‘Ž sub four, and π‘Ž sub five. We know that the sequence will start with an index 𝑛 of one because we’re told that 𝑛 is greater than or equal to one.

So let’s use this formula and say, for example, that we wanted to work out the third term, π‘Ž sub three. We can use the formula to tell us that π‘Ž sub three is one-quarter of π‘Ž sub two; it’s one-quarter of the second term. But the problem is, is that we don’t yet know the second term of the sequence. We could work out the second term as one-quarter of the first term, but what is the first term?

Well, we’re told that π‘Ž sub one is negative 27. In this sort of formula, which is a recursive formula, then we need to be given at least one of the terms in order to give us somewhere to start with the sequence. Looking at the second term, as previously mentioned, we can find this by taking one-quarter of the first term. We can work out one-quarter multiplied by negative 27. And negative 27 over four is the simplest form of this fraction, and so that’s the second term.

The third term π‘Ž sub three is one-quarter multiplied by the second term, which is one-quarter times negative 27 over four. So we have negative 27 over 16. The fourth term is one-quarter times the third term of negative 27 over 16, which is negative 27 over 64. Finally, the fifth term is one-quarter times negative 27 over 64, which is negative 27 over 256. We can then give the answer that the first five terms of the given sequence are negative 27, negative 27 over four, negative 27 over 16, negative 27 over 64, and negative 27 over 256.

In the final example, we will consider what the graph of a geometric sequence would look like.

True or False: The terms of a geometric sequence can be plotted as a set of collinear points.

Let’s begin this question by recalling that a geometric sequence is a sequence which has a common ratio between any two consecutive terms. In order to consider whether a geometric sequence might be collinear, which means lying on a straight line, then it might be useful to take a few examples of geometric sequences.

So let’s take the sequence one, three, nine, 27, and so on. We can say that it’s geometric because the common ratio is three. Any term is found by multiplying the previous term by three. If we were to graph these, then we’d be graphing the 𝑛 or the index value alongside the term value. We could start with the coordinate one, one because the first term has a value of one. The second coordinate would be that of two, three. The term with index two has a value of three.

However, a third coordinate of three, nine might begin to reveal the pattern in this geometric sequence. We would not have a straight line. In fact, what we would have would be an exponential graph.

So let’s try another geometric sequence. This time, let’s try a decreasing geometric sequence. The sequence negative two, negative four, negative eight, negative 16, and so on has a common ratio of two. Let’s try plotting these term values. Once again, we can see that these points would not lie on a straight line.

But there is another type of geometric sequence, and that’s an alternating sequence. The signs of the terms alternate between positive and negative. This is because the ratio is a negative value. When we plot this geometric sequence, we get a graph that looks like this. So far, none of the sequences that we have considered have created a set of collinear points. So let’s consider what type of sequence would.

Well, if we had a sequence which produces a straight line, that means that when the index increases, the terms increase by a constant or a constant difference. This type of sequence is actually an arithmetic sequence. And it’s defined by a common difference between any two consecutive terms. Remember that a geometric sequence has a common ratio between terms, and that common ratio cannot be equal to one. It is therefore not possible that a geometric sequence can be plotted as a set of colinear points. That means that the statement in the question is false.

It’s worth noting that arithmetic sequences are always linear but geometric sequences are never linear. In fact, they would create an exponential function.

We can now summarize the key points of this video. We began by considering that a geometric sequence is a sequence of nonzero numbers with a common ratio π‘Ÿ, which is not equal to one, between any two consecutive terms in the sequence. We saw how we can calculate this common ratio π‘Ÿ by taking any term in the sequence and dividing it by the term beforehand. This can be written as π‘Ÿ equals π‘Ž sub 𝑛 plus one over π‘Ž sub 𝑛.

We also saw that a given geometric sequence may be defined as a set of numbers π‘Ž sub one, π‘Ž sub two, π‘Ž sub three; a recursive formula; or an explicit formula. Finally, we saw that geometric sequences can be increasing, decreasing, or alternating.

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