Video: Introducing Geometric Sequences

Defining, identifying, and exploring geometric sequences through a series of examples. Learn how to find the common ratio between terms (positive, negative, and rational examples) and use it to produce a general formula for the 𝑛th term of the sequence.

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Video Transcript

In this video we’re gonna look at geometric sequences and we’re gonna see how to write down a general formula for a particular geometric sequence. Then we’ll go on and answer a few typical questions. So let’s start off with the definition. A sequence of numbers are called a geometric sequence if each term is multiplied by the same common ratio to get the next term. So for example, we’ve got a sequence of numbers three, six, 12, 24, and so on. And in this case, three is our first term. And in each case, to get the next number in the sequence, we’re simply doubling each term. So the common ratio is two. To get one term, we simply multiply the previous term by two.

In another example, the sequence 10, 15, 22.5, 33.75, and so on, the first term in the sequence is 10. And I need to multiply each term in the sequence by 1.5 to get the next term. So my common ratio is 1.5. Another example would be the sequence seven, seven-tenths, seven hundredths, seven thousandths, and so on. Now for this sequence, my first term will be seven. And I need to multiply each term by a tenth to get the next term. So my common ratio is a tenth. And another example would be the sequence 32, negative 16, eight, negative four. And in this case, the first term will be 32. And I’d need to multiply each term by negative a half to get the next term. So my common ratio would be negative a half. So my common ratio then could be positive, could be negative, could be integer, fraction, decimal. There’s all sorts of possibilities for what that might be.

Now depending on where you live, there are several different styles of notation that are commonly used for geometric sequences. For example, some people call the first term π‘Ž or π‘Ž one or 𝑒 one or 𝑑 in parentheses one, or 𝑑 one. But in this video, I’m gonna use the convention π‘Ž one is the first term. And likewise, there are a different ways that people express the 𝑛th term. It could be π‘Ž 𝑛; it could be 𝑒 𝑛, 𝑑 𝑛, or 𝑑 (𝑛) written like this. I’m gonna be using this one in this video. But luckily, most people seem to use π‘Ÿ to represent the common ratio. So again that’s what I’ll use.

So let’s write down the first five terms of a geometric sequence with the first term, π‘Ž one is equal to 12, and a common ratio, π‘Ÿ is equal to a third.

Well, we’re told that the first term is 12. So let’s just write that down; the first term is 12. Now, the common ratio is a third; that means I have to multiply each term by a third in order to get the next term. So to get the second term, I need to multiply 12 by a third. And a third of 12 is four. And to get the third term, I need to multiply the second term by a third. So that’s four times a third which is four over three, four-thirds. And to get the fourth term, I’m going to need to multiply that by a third. And four-thirds times a third is four-ninths. So my fourth term is four-ninths. And then let’s do that one more time. Four-ninths times a third is four twenty-sevenths. And that process carries on forever. That geometric sequence carries on for as many times as you’d like to write down.

Just going over the notation a bit more, remember π‘Ž one, the first term, is twelve; π‘Ž two, the second term, is four; π‘Ž three, the third term, is four-thirds; π‘Ž four, the fourth term, is four-ninths; π‘Ž five, the fifth term, is four twenty-sevenths; and so on. Now in order to get the second term, we took the first term and multiplied it by the common ratio. And to get the third term, we took the second term and multiplied it by the common ratio. So π‘Ž three is π‘Ž two times π‘Ÿ. But remember, π‘Ž two is π‘Ž one times π‘Ÿ. So I can replace π‘Ž two in here with π‘Ž one times π‘Ÿ. So π‘Ž two is π‘Ž one times π‘Ÿ. And then we multiply that by π‘Ÿ to get π‘Ž three.

Now to get π‘Ž four, we took π‘Ž three and multiplied that by the common ratio. So remember π‘Ž three is π‘Ž one times π‘Ÿ times π‘Ÿ. So we multiplied that by π‘Ÿ to get π‘Ž four. And likewise we multiply that by π‘Ÿ again to get the next term. So π‘Ž five is π‘Ž one times π‘Ÿ times π‘Ÿ times π‘Ÿ times π‘Ÿ. Now, we can write that in power format. So instead of writing π‘Ÿ times π‘Ÿ times π‘Ÿ times π‘Ÿ, we can write π‘Ÿ to the power of four. So π‘Ž one is just twelve or π‘Ž one, π‘Ž two is π‘Ž one times π‘Ÿ to the power of one, π‘Ž three is π‘Ž one times π‘Ÿ to the power of two, π‘Ž four is π‘Ž one times π‘Ÿ to the power of three, and so on.

So I’m just gonna complete that sequence by saying π‘Ž one is just π‘Ž one times one. But instead of just writing one, I’m going to say π‘Ÿ to the power of zero. Remember anything to the power of zero is one. So now we got a bit of a pattern. The first term is just the first term times π‘Ÿ to the power of zero. The second term is the first term times π‘Ÿ to the power of one. The third term is the first term times π‘Ÿ to the power of two. The fourth term is the first term times π‘Ÿ to the power of three. And the fifth term is the first term times π‘Ÿ to the power of four. So the power of π‘Ÿ is one less than the position of the term in the sequence.

So if I say 𝑛 is the position in the sequence, the 𝑛th term, π‘Ž 𝑛, is simply the first term times π‘Ÿ to the power of one less than 𝑛. So that’s π‘Ž 𝑛 equals π‘Ž one times π‘Ÿ to the power of 𝑛 minus one, one less than 𝑛. So now, we’ve got a nice little formula that tells us the any-any term in the sequence. So we don’t have to keep multiplying by a third. We can go straight to that term in the sequence.

So let’s see an example of that.

So let’s use the general formula to find the value of the seventh term in this particular sequence.

Well if the general term π‘Ž 𝑛 is equal to π‘Ž one times π‘Ÿ to the power of 𝑛 minus one. That’s the first term time the common ratio- times the common ratio to the power of 𝑛 minus one. Well, our first term was 12; we were told in the question. So π‘Ž one is 12. And our common ratio was a third. So our general formula for this particular sequence is π‘Ž to the 𝑛 is equal to 12 times a third to the power of 𝑛 minus one. So to find the seventh term, we put 𝑛 is equal to seven. And π‘Ž seven, the seventh term, is gonna be twelve times a third to the power of seven minus one. Well seven minus one is six. So the seventh term is going to be 12 times a third to the power of six. And a third to the power of six is one over 729. So it’s gonna be 12 over one over 729, which is four over 243. Let’s have a look at another question.

Write down the first term and the common ratio for the following geometric sequence: 10, negative five, five over two, negative five over four, and so on.

Well clearly, the first term is equal to 10. So π‘Ž one is equal to 10; that bit was quite easy. And the common ratio is what do we multiply each term by to get the next term. So I’m just gonna label all of my terms π‘Ž one, π‘Ž two, π‘Ž three, and π‘Ž four, and so on. And then I’m just going to write a little formula for how do I get from one term to the next term. Well, if I multiply the first term by the common ratio, π‘Ÿ, I get the second term. If I multiply the second term by the common ratio, π‘Ÿ, I get the third term. If I multiply the third term by the common ratio, π‘Ÿ, I get the fourth term, and so on and so on and so on. So looking at that first equation, if I divide both sides of the equation by π‘Ž one, I get that π‘Ÿ is equal to π‘Ž two over π‘Ž one. Now, if I divide both sides of the second equation by π‘Ž two, I get that π‘Ÿ is equal to π‘Ž three over π‘Ž two and similarly for the third equation.

So to work out the value of π‘Ÿ, I just take the value of one term and divide it by the value of the previous term. Now remember in a geometric sequence, it’s a common ratio. So it doesn’t matter whether I take the second and the first or the third and the second or the fourth and the third. As long as I take two consecutive terms, I will always find the same answer for π‘Ÿ. Well looking at these numbers here, the easiest pair to use is gonna be π‘Ž one and π‘Ž two. So π‘Ÿ is equal to π‘Ž two divided by π‘Ž one. π‘Ž two is negative five and π‘Ž one is ten. So the common ratio is negative five divided by 10 and that simplifies to negative a half.

So to get from each term to the next term, I have to multiply by negative a half. 10 times negative a half is minus five, negative five times negative a half is five over two, and so on. So these two facts here, π‘Ž one equals 10 and π‘Ÿ equals negative a half, uniquely defines this sequence. When we know this, we can generate all the terms in the sequence if we’re prepared to put in enough time and doing enough multiplying.

Okay, let’s see some sequences and try and work out whether or not they are geometric.

Well in fact, this question says, is the following sequence arithmetic or geometric?

Now, if you remember, an arithmetic sequence is one in which each term has a common difference added in order to generate the next term. So to answer this question, we just need to see what do we add to get from one term to the next and see if that is constant and what do we multiply to get from one term to the next and see if that is constant. So with this particular set of numbers, if I add one every time, I’m generating that sequence. But to multiply one term to get the next term, the ratio keeps changing; it’s not a common ratio. So the fact that we got a common difference means that this is an arithmetic sequence.

Let’s try this one then.

Is the following sequence arithmetic or geometric? 11, 33, 99, 297, and so on.

Well, if we were adding we would get- have to add different numbers each time to get the next number in the sequence. But if we multiply each term by three, we generate the next term. So we’ve got a common ratio of three. So the fact that we’ve got a common ratio rather than a common difference tells us that we got a geometric sequence.

Now this one, is the following sequence arithmetic or geometric? One, two, four, seven, 11, and so on. Well, I have to add a different number each time to generate the next term. So it’s not an arithmetic sequence. And I need to double the first to get the second and double the second to get the third. But after that I’m not doubling; I am having to multiply by different numbers. So there is not a common difference and there is not a common ratio. So it’s neither arithmetic or geometric. So an interesting sequence, but it’s not arithmetic or geometric.

Okay one last one of these then.

Is the following sequence arithmetic or geometric? 5.2, 5.2, 5.2, 5.2, and so on.

Well, what do you think? Is it arithmetic or is it geometric? What do I have to add in order to get from one term to the next? Well, it’s nothing in each case. I’m adding zero. So I’ve got a common difference of zero. Well that’s a bit weird. But that is an arithmetic sequence. And what do I have to multiply each term by to get the next term? Well in each case, I’m just multiplying by one. So again it’s a bit of a weird sequence. But it is a geometric sequence because we got a common ratio of one. So it’s a pretty weird example I’ll grant you that. But it is both arithmetic and geometric if we follow those strict rules. The common difference is zero and the common ratio is one.

Now, let’s look at another question.

Find the next three terms in the geometric sequence 100, negative 10, one, negative 0.1, 0.01, and so on. So we’ve got the first five terms, π‘Ž one, π‘Ž two, π‘Ž three, π‘Ž four, and π‘Ž five. And the first thing I need to do is to work out what the common ratio is. What do I need to multiply π‘Ž one by to get π‘Ž two and so on? And if you remember the way that we do this is we divide one term by its previous term to find out what the ratio is. And looking through there, I reckon the second and third terms are gonna be the easiest ones to divide. And although you get the same answer no matter which consecutive pair you divided, this one’s easy because it’s one divided by negative 10 which is negative a tenth.

So the common ratio is negative a tenth. We need to multiply each term by negative a tenth to get the next term. So to find the next three terms, I just need to take the last term that we had and multiply that by negative a tenth, then multiply that by negative a tenth, and multiply that by negative a tenth. So the sixth term is the fifth term times negative a tenth; that’s, 0.01 times negative a tenth, which is negative 0.01. So multiplying by negative a tenth is the same because it’s dividing by negative 10. So this is relatively easy to do. So the sixth term times negative a tenth, the two negatives are gonna cancel out to make it positive. And 0.001 divided by 10 is 0.0001. And doing the same, the eighth term is negative 0.00001. So we just need to write our answer out there nice and clearly.

Now, we talked a bit about finding a formula for a general term. So here’s a question. So let’s have a look at this.

Find a formula for the general term of the geometric sequence three, 15, 75, 375, 1875.

Well, our first term is three. So that bit is easy and I’ve got to work out what the common ratio. And remember, we’re just gonna do a division of one term divided by its previous term. And the easiest numbers to work with here I think are gonna be these two, π‘Ž one is three and π‘Ž two is 15. So the common ratio is π‘Ž two divided by π‘Ž one which is 15 over three which is five. Now remember, we were told in the question that this is a geometric sequence. So it didn’t matter which pair of terms β€” consecutive terms β€” that we chose; we would have got the same answer π‘Ÿ equals five. But just by choosing these first two terms, the numbers were simpler.

So we know that π‘Ž one, the first term, is three and the common ratio is five. So we can put that into our formula. And remember, to work out the value of any particular term in the sequence, what we do is we take the first term and we’re gonna keep multiplying it by the common ratio. Now, what we have to do if we are looking for the fifth term, we’ve only had to multiply that first term by the common ratio four times in order to get that. So whatever term we’re looking for, it’s the common ratio to the power of that term minus one. And we just worked out that π‘Ž one was three and π‘Ÿ is five. So to work out the value of term 𝑛 in this particular sequence, it’s gonna be three times five to the power of whatever term that is minus one, 𝑛 minus one.

So let’s take that one little step further now.

And we’ve got to find a formula for the general term of the geometric sequence negative 512, 128, negative 32, eight, negative two, and so on. And we gotta use that formula to find the value of the twelfth term of the sequence.

Or we can just read off the first term there, negative 512. And now we’ve gotta work out the common ratio. So it’s any pair of consecutive terms, one divided by the other, the second divided by the first. So I’m gonna choose π‘Ž four and π‘Ž five in this case cause they look like the easiest numbers to work with. So π‘Ž five is negative two and π‘Ž four is eight. So the common ratio is gonna be negative two over eight, which is negative a quarter. So now, I’ve got those two important bits of information. It’s pretty easy to work out the general formula. So remembering π‘Ž 𝑛 is equal to π‘Ž one times π‘Ÿ to the 𝑛 minus one, let’s substitute the values 𝑛 for π‘Ž, π‘Ž one and π‘Ÿ.

So my general formula is π‘Ž 𝑛, the nth term is equal to negative 512 times negative a quarter to the power of 𝑛 minus one. So I’m trying to find now the twelfth term. So 𝑛 equals 12, which means that the twelfth term, π‘Ž 12, is equal to negative 512 times negative a quarter to the power of 12 minus one. Well, 12 minus one is eleven. And when I work all that out, I’ve got π‘Ž 12 is equal to one over 8192.

So let’s quickly summarise what we’ve looked at there. The geometric sequence is where you multiply each term by a common ratio to get the next term. For example, three, six, 12, 24. I double each term to get out the next term. And in this case, the common ratio is two and the first term was three. To work out the common ratio, which we’ve called π‘Ÿ, you just take a term and divide it by the previous term. In general, the 𝑛th term is simply the first term multiplied by π‘Ÿ and minus one times. So in the case of our example, our 𝑛th term will be three times two to the power of 𝑛 minus one.

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