Lesson Video: Finding the 𝑛th Term of a Geometric Sequence | Nagwa Lesson Video: Finding the 𝑛th Term of a Geometric Sequence | Nagwa

Lesson Video: Finding the 𝑛th Term of a Geometric Sequence Mathematics • Second Year of Secondary School

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In this video, we will learn how to write explicit and recursive formulas for geometric sequences to find the value of the nth term in a geometric sequence and how to find a term’s order given its value.

16:45

Video Transcript

In this video, we will learn about geometric sequences. We will discover how to find the general or 𝑛th term and how to find the term-to-term rule. We’ll also cover how to find the order of a term given its value.

The first thing to note about a geometric sequence is these are sequences where the ratio between terms is constant. Notice that that’s different to arithmetic sequences, where it’s the difference between terms that’s constant. We can describe a geometric sequence in one of two ways, either by using a term-to-term rule or by using a position-to-term rule. The position-to-term rule is often called the 𝑛th term, and it’s very helpful for finding the value of a specific term. For example, if we wanted to find the 15th term of a sequence, we could directly evaluate it by substituting 15 into the 𝑛th term rule rather than having to work out all the terms up to the 15th one using the term-to-term rule.

So, let’s think about some of the notation that we use in geometric sequences. We say that if the first term is denoted as the letter π‘Ž and the common ratio is π‘Ÿ, then our sequence would look like this. The first term is π‘Ž. The second term would be π‘Žπ‘Ÿ because we’ve multiplied the first term π‘Ž by the common ratio π‘Ÿ. Multiplying the second term π‘Žπ‘Ÿ with another π‘Ÿ would give us π‘Žπ‘Ÿ squared. We can signify the terms by using subscript notation. For example, the first term would be written as π‘Ž sub one, the second term would be written as π‘Ž sub two, the third as π‘Ž sub three, and so on.

So, how could we find a rule to find the 𝑛th term which is written as π‘Ž sub 𝑛? Well, we could start by noticing that each term has an exponent value of π‘Ÿ which is one less than the term number. We know that the 𝑛th term would still have an π‘Ž-value, and the exponent of π‘Ÿ would be one less than 𝑛. We can, therefore, say that the 𝑛th term of any geometric sequence can be written as π‘Ž times π‘Ÿ to the power of 𝑛 minus one. We remember, of course, that it’s just the π‘Ÿ that’s taken to the power of 𝑛 minus one and it doesn’t include the π‘Ž as well. So, let’s see how we can put this formula into practice to find the 𝑛th term of our first geometric sequence.

Find the general term of the geometric sequence negative 76, negative 38, negative 19, negative 19 over two.

Another way of phrasing general term is the 𝑛th term. So, we’re looking for the 𝑛th term of this geometric sequence, which is a sequence which has a common ratio between the terms. So, let’s have a look at the sequence and see what we can determine. Firstly, we can see the first term in the sequence is negative 76. When we’re working with geometric sequences, we usually use the letter π‘Ž to signify the first term of the sequence. In order to find the 𝑛th term, we’ll also need to find π‘Ÿ, the common ratio. When we consider a general geometric sequence written as π‘Ž then π‘Žπ‘Ÿ then π‘Žπ‘Ÿ squared, and so on, we can find the common ratio π‘Ÿ by dividing any term by the immediately preceding term.

So, here, we could take the second term of negative 38 and divide it by negative 76. Therefore, π‘Ÿ is equal to one-half. Notice that even if we’d taken two different terms, for example, if we divided the third term of negative 19 by the second term of negative 38, we would still have found the common ratio π‘Ÿ to be one-half. After all, if it wasn’t the same, then we wouldn’t have a geometric sequence.

So now that we found the values of π‘Ž and π‘Ÿ, we recall the general formula for the nth term of a geometric sequence. π‘Ž sub 𝑛, that’s the 𝑛th term, is equal to π‘Ž times π‘Ÿ to the power of 𝑛 minus one. All we need to do now is plug in the values of π‘Ž equals negative six and π‘Ÿ equals a half into this formula. This gives us π‘Ž sub 𝑛 equals negative 76 times a half to the power of 𝑛 minus one. As we can’t simplify this any further, then that’s our answer for the general term or the 𝑛th term of the geometric sequence.

Let’s have a look at another question where we’re finding the 𝑛th term of a slightly more complex sequence.

Find, in terms of 𝑛, the general term of the sequence one-fourth, nine over 16, 81 over 64, 729 over 256, and so on.

In this question, we’re given the first four terms of this sequence. There doesn’t look like there’s a common difference between the terms, so we could say that this is definitely not an arithmetic sequence. We can check if it’s a geometric sequence which would have a common ratio between the terms by seeing if we can work out what that common ratio would be.

In order to find the ratio π‘Ÿ between the first two terms, we would take the second term, nine over 16, and divide by the term before it, one-quarter. We can recall that to divide by one-quarter, that would be equivalent to multiplying by the reciprocal, which would be four over one. We can simplify the four on the numerator and the 16 on the denominator by taking out a factor of four. We then multiply our numerators and denominators. Nine times one would give us nine, and four times one gives us four. We can check if there’s the same ratio between the third term and the second term. So, we calculate 81 over 64 divided by nine over 16. We can see that the reciprocal of nine over 16 would be 16 over nine. And we simplify the fractions before multiplying the numerators and denominators, which gives us the same common ratio π‘Ÿ of nine over four. And if we check this by working out what we need to multiply the third term 81 over 64 by in order to get the fourth term of 729 over 256, we get the same answer of nine over four.

So, let’s think about how we would find the general term of this sequence. We can recall that the general term is another way of saying the 𝑛th term. We can recall that the formula to find the 𝑛th term π‘Ž sub 𝑛 is that π‘Ž sub 𝑛 is equal to π‘Ž times π‘Ÿ to the power of 𝑛 minus one. The π‘Ÿ-value represents the common ratio, and the π‘Ž-value represents the first term of the sequence. We’ve already established that π‘Ÿ is equal to nine over four, and the π‘Ž-value, the first term in the sequence, would be one-quarter. We can then take the values of π‘Ž and π‘Ÿ and plug them into this formula. This gives us π‘Ž sub 𝑛 equals one-quarter times nine-quarters to the power of 𝑛 minus one.

While this is a perfectly valid answer for the 𝑛th term of the sequence, it might be worth seeing if we can simplify this right-hand side any further. We can start our simplification by thinking about what happens when we have a power of a fraction. A fraction with a certain exponent value is equivalent to the numerator with that exponent over the denominator with that exponent. If we then consider these fractions multiplied, multiplying the numerators one times nine to the power of 𝑛 minus one would give us the value on the numerator of nine to the power of 𝑛 minus one. Multiplying the denominators would give us four times four to the power of 𝑛 minus one.

If we look at the denominator, we can apply another exponent rule that π‘₯ to the power of π‘Ž multiplied by π‘₯ to the power of 𝑏 is equal to π‘₯ to the power of π‘Ž plus 𝑏. On the denominator, the π‘₯-value here would be four. And our π‘Ž- and 𝑏-values, that’s the exponents, would be represented by one and 𝑛 minus one. Adding one and 𝑛 minus one would leave us with the value of 𝑛. So, we’ve found a fully simplified answer for the general term of this sequence in terms of 𝑛. It’s nine to the power of 𝑛 minus one over four to the power of 𝑛.

In the two questions that we’ve just seen, we’ve been finding the 𝑛th term or the position-to-term rule. In the next question, we’ll have a look at how we would find a term-to-term rule.

Find a recursive formula for the sequence 486, 162, 54, 18, six, two, two-thirds.

So here we have a sequence which begins with the number 486, and we’re asked to find a recursive formula. This means that instead of finding a general term or an 𝑛th term for the sequence, we’re going to find a term-to-term rule. It means we’ll be thinking about how we go from one term to another term. Let’s begin by seeing if we can find a common ratio between the terms. We can signify this with the letter π‘Ÿ. In a geometric sequence, we say that the first term is π‘Ž, the second term would be π‘Ž times the common ratio π‘Ÿ, the third term would be π‘Žπ‘Ÿ squared, and the fourth term would be π‘Žπ‘Ÿ cubed, and so on. Therefore, to find the ratio between terms, we take any term and divide by the term immediately before it.

So, we could, for example, take the second term of 162 and divide by the term before it, 486. Or if it’s a geometric sequence, as we assume, then we could take the third term and divide by the second term. We could even take the sixth term of two and divide by the fifth term. We should hopefully see straightaway that two-sixths can simplify to one-third. So, do the other two fractions here also simplify to one-third? Well, if we take 54 and add 54, that’s 108. And adding another 54 would give us 162. So, this fraction would also simplify to one-third. 162 over 486 would also give us one-third.

It’s looking like we have a common ratio of one-third. In fact, if we take any two consecutive terms in this sequence, we find that we need to multiply the first one by one-third to get the second. The terms we have been given definitely do have a common ratio of one-third. So, does this answer the question then to find a recursive formula? Is it sufficient to say we would multiply a term by one-third to find the next one?

Well, not quite. We can use some more formal notation here. We say that the first term π‘Ž one will be equal to 486. To find the second term, we know that that would be 486 times the ratio π‘Ÿ of one-third. So, π‘Ž sub two, the second term, is equal to one-third of π‘Ž sub one. In the same way, if we wanted to find the third term and we didn’t know the value of it, we take the second term and multiply it by the ratio π‘Ÿ of one-third. We could continue like this to say the fourth term would be equal to one-third times the third term.

So, if we wanted to find the 𝑛th term of this sequence, we would calculate one-third multiplied by the term before, written as π‘Ž sub 𝑛 minus one. This is where we get the helpful formula from, that if we want to find the 𝑛 plus oneth term of a sequence, then we multiply the common ratio by the 𝑛th term. It sort of looks like the formula for the 𝑛th term of a sequence, but this time the term is based on the term before it rather than the starting term of a sequence. But here, we can give our answer that any term in the sequence π‘Ž sub 𝑛 can be calculated by a third times the term before it, π‘Ž sub 𝑛 minus one.

In the next question, we’ll see how we can find the order of a term given its value and the 𝑛th term of the sequence.

Find the order of the term whose value is 4374 in the geometric sequence π‘Ž sub 𝑛 equals two-thirds times three to the power of 𝑛.

Let’s start by looking at the information that we’re given. This value of π‘Ž sub 𝑛 represents the 𝑛th term of this sequence. We’re asked to find the order of the term whose value is 4374. That means that we’ve got a sequence, and somewhere in this sequence is this value of 4374. The order of this term means we’re really asking, is it the second term, the 10th term, the 100th term? That’s what we need to find out. We can do this by saying let’s make the order of this term 𝑛, and then our 𝑛th term will be 4374. We could then fill this into the formula and rearrange to find this value of 𝑛, which would give us the order of this term.

We can start our rearranging by dividing both sides of this equation by two-thirds. On the left-hand side, we can recall that to divide by a fraction, we multiply by its reciprocal. And on the right-hand side, we’ll be left with three to the power of 𝑛. We can simplify the values on the left-hand side. So, we work out 2187 multiplied by three, which gives us 6561 is equal to three to the power of 𝑛.

Now, at this stage, there’s a branch of mathematics called logarithms, which would help us solve this problem directly. But as most people learn this long after they learn about geometric sequences, we’ll use a bit of trial and improvement here instead. Remember that a value like three to the power of 𝑛 equals 6561 is really equivalent to saying three to the power of what gives us this value. You might know your first powers of three off by heart, up to roughly three to the power of four equals 81. We could then continue with a few more by multiplying each of the values by three as we go up. If we’re using a non-calculator method, we’ll probably need to start using some pencil and paper working out. But then, we find that three to the power of eight is equal to 6561. This means that our 𝑛-value here must be equal to eight. So, we can give our answer that the order of the term whose value is 4374 is eight, as it would be the eighth term in this sequence.

We can now summarize what we’ve learnt in this video. Firstly, we saw that a geometric sequence is a sequence where the ratio between terms is constant. For a first term π‘Ž and a common ratio π‘Ÿ, we have the terms of the sequence as π‘Ž, π‘Žπ‘Ÿ, π‘Žπ‘Ÿ squared, π‘Žπ‘Ÿ to the third power, and so on. To find the common ratio π‘Ÿ given a value of terms in a sequence, we can divide any term by its preceding term. For example, π‘Ÿ can be found by dividing the third term by the second term. The position-to-term rule or the 𝑛th term can be written as π‘Ž sub 𝑛 equals π‘Ž times π‘Ÿ to the power of 𝑛 minus one.

Finally, we saw the term-to-term rule, or recursive formula, which can be written as π‘Ž sub 𝑛 plus one equals π‘Ÿ times π‘Ž sub 𝑛, remembering that this means if we want to find the value of the term with order 𝑛 plus one, then we take the value of the term with order 𝑛 and multiply it by the common ratio π‘Ÿ.

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