# Video: Further Geometric Sequences

Tim Burnham

This video explains how to tackle trickier geometric sequence questions for which you are not given the value of the first term or for the common value of which there are multiple possible values. We also cover geometric sequences represented with a recursive formula.

17:35

### Video Transcript

In this video, we’re gonna look at some slightly trickier geometric sequence questions than we looked at in the introduction video. And we’ll also explain recursive sequence formulae. Then we’ll see some real-world applications of geometric sequences in action.

Now remember first, a geometric sequence is a sequence of numbers where each term could be multiplied by a common ratio to get the next term. For example, the sequence three, six, twelve, twenty-four, and so on. Our first term is three, and we have to multiply each term by two in order to generate the next term. So our common ratio is two. And if we use the terminology 𝑎 one for the first term and 𝑟 for the common ratio, then 𝑎 𝑛 for the 𝑛th term, then we can use this general formula to generate the 𝑛th term. The 𝑛th term, 𝑎 𝑛, is equal to the first term, 𝑎 one, times the common ratio, 𝑟, to the power of 𝑛 minus one. And to work out the value of 𝑟, we just take any term and divide it by the previous term. For example, take the second term, divide it by the first term, that will give you the common ratio. Or the third divided by the second, or the fourth divided by the third, and so on.

And this general formula for the 𝑛th term gives you a way of calculating the value of a specific term directly. For example, if the first term is equal to nine, and the common ratio was two, and we want to find the twelfth term, then we can start by writing out the formula for the 𝑛th term, it’s the first term nine times the common ratio two to the power of 𝑛 minus one. Now we’re looking for the specific case where 𝑛 is equal to twelve to find the twelfth term. That means that the twelfth term is equal to nine times two to the power of twelve minus one. In other words, nine times two to the power of eleven, which is eighteen thousand four hundred and thirty-two.

But there’s also another sort of formula that we can write down for a geometric sequence and that’s in the recursive formula format. Now put simply, this is a formula that just tells you what you need to do to one term to get to the next. Now that’s not too tricky because you know in a geometric sequence that you just need to multiply each term by the common ratio to get the next term. So the second term is just the first term times 𝑟, the common ratio. And the third term is just the second term times the common ratio, and so on.

So how can we generalize this into a formula? Well the 𝑛th term, 𝑎 𝑛, is simply equal to the previous term, the 𝑛 minus oneth term, 𝑎 𝑛 minus one, times the common ratio 𝑟. So there it is, the recursive formula for a geometric sequence. But that’s just the term-to-term rule. We also need to specify a start point for the sequence. Otherwise, we wouldn’t uniquely determine that sequence. So for example, here’s a geometric sequence. This is the term-to-term rule, this is telling you that we’re multiplying each term by two to get the next term, and this tells you to start at nine. So you’ve got a unique set of numbers, nine, then times two, then times two, then times two, and so on.

In fact, it doesn’t have to be the very first term that you specify. If you specify any term, you can work out the entire sequence from that. So for example, if we said that the term-to-term rule was that we have to multiply each term by two to get the next term, and that the fifth term was thirty-two, then we can work out the sixth term by just multiplying the fifth term by two. So the sixth term is sixty-four, and the seventh term is a hundred and twenty-eight, and so on. But we could also go the other way because we know that the fifth term is the fourth term multiplied by two, and we know that the fifth term was thirty-two, so 𝑎 four times two is equal to thirty-two. Now if I divide both sides of that equation by two, I get 𝑎 four is equal to sixteen. And I can keep following that method through, the-the fourth term is just the third term multiplied by two. And we’ve just worked out that that is sixteen. So if I divide both sides of that by two, we know that 𝑎 three is eight, and so on, and so on.

Okay then, let’s see a question that involves using the recursive formula. The geometric sequence has the recursive formula 𝑎 𝑛 is equal to negative a half of 𝑎 𝑛 minus one, and its first term is sixty-four. Find the general formula for the 𝑛th term of the sequence and use it to find the twentieth term.

So we know that 𝑎 one, the first term, is sixty-four. It told us that in the question. And the common ratio was, at the recursive formula is telling us that we multiply each term by negative a half in order to get the next term, so our common ratio is negative a half. And we know that the general form of this formula for the 𝑛th term is, the 𝑛th term is equal to the first term times the common ratio to the power of 𝑛 minus one. And now we know the values of 𝑎 one and 𝑟. It’s relatively straightforward to plug those in there, to get our general formula. And to find the twentieth term, we just need to put 𝑛 equals to twenty. So the twentieth term 𝑎 twenty is equal to sixty-four times negative a half to the power of twenty minus one, which is sixty-four times negative a half to the power of nineteen of course, which is sixty-four times negative one over five hundred and twenty-four thousand two hundred and eighty-eight. And when you work that out, you get minus one over eight thousand one hundred and ninety-two. So although the term recursive formula is a different bit of terminology, it’s relatively easy to use once you know how to use it.

Now another bit of terminology that you might encounter is the term geometric series. Now we’re gonna talk about that in another video. But just so that you know what it is if you hear the term, it’s the sum of the terms in a geometric sequence. So when we use the term geometric sequence, we mean the set of numbers that make up that sequence. So in this case, it’s a list of five, then ten, then twenty, then forty, and eighty, and so on. But if we were to use the phrase geometric series, we’d mean five plus ten plus twenty plus forty plus eighty. So it’s the sum of the numbers in that sequence, so plus all the other numbers as well of course. So we’re not really gonna do anything with this, but I just wanted you to be aware of what it meant, in- just in case you encountered that particular phrase.

Let’s have a look at some of these slightly trickier geometric sequence questions then. The fourth and fifth terms in a geometric sequence are one hundred and twenty-five and six hundred and twenty-five. Find the first, second, third, and tenth terms in that sequence.

So the question tells us that the fourth term is a hundred and twenty-five and the fifth term is six hundred and twenty-five. Now we can put those two bits of information together to find out the common ratio. All we have to do is take the value of one term and divide by the previous term, and that’ll tell you what the common ratio is. So the fifth term six hundred and twenty-five, fourth term a hundred and twenty-five, six hundred and twenty-five divided by a hundred and twenty-five is five. What that’s telling us is, we’d need to multiply a hundred and twenty-five by five to get the next term six hundred and twenty-five. Now we know that the fourth term is simply the third term times the common ratio. And we know what the fourth term is, it’s a hundred and twenty-five. So a hundred and twenty-five is equal to the third term times five. Now if I divide both sides of that equation by five, I find out that the third term is equal to twenty-five. And again, following the pattern, the third term is just the second term times five. I know what the third term is now, I just worked it out, twenty-five, so twenty-five is equal to the second term times five. Now I need to divide both sides of that equation by five. And I can see that the second term is five. So then repeating that again, I find out that the first term was one. So I’ve got my first, second, and third terms there. Now I just need to work out what the tenth term is. Well I could go through and then just keep multiplying by five to work it out that way, or I could work out the general formula and then plug in 𝑎 ten to that and find the answer.

Now the general format of the general formula for the 𝑛th term is, the 𝑛th term is equal to the first term times the common ratio to the power of 𝑛 minus one. Now I know what the common ratio and the first term are. I can put them into that formula so the 𝑛th term is one times five to the power of 𝑛 minus one. Well I don’t really need to write the one times, so this is my formula here 𝑎 𝑛, the 𝑛th term, is equal to five to the power of 𝑛 minus one. Now I want to find out the tenth term, so 𝑛 is equal to ten. The tenth term is equal to five to the power of ten minus one, which is five to the power of nine. And just calculating that, I get 𝑎 ten, the tenth term, is equal to one million nine hundred and fifty-three thousand one hundred and twenty-five.

And our next question is: Find the general formula and the eleventh term of the geometric sequence with a third term of fifty-four and a fifth term of six. So this is definitely a bit more tricky.

So I can extract from the question, the third term is fifty-four, so 𝑎 three is fifty-four. The fifth term is six, so 𝑎 five is six. But I don’t know what 𝑎 four is, and I don’t yet know what the ratio is. But I do know that the ratio is one term divided by the previous term. So the second term divided by the first term, or the third term divided by the second term, and so on. Now I know what the third and the fifth terms are. So looking at these two things here, I know that and I know that, this is one unknown, so I could-I could put an equation together involving 𝑎 three, a four, and 𝑎 five in this kind of format. So I know 𝑎 four divided by 𝑎 three, which is fifty-four, is equal to 𝑎 five, which is six, divided by 𝑎 four. Now I could multiply both sides by fifty-four, so that the fifty-fours cancel on the left. And that gives me three hundred and twenty-four over 𝑎 four is equal to 𝑎 four. Now if I multiply both sides by 𝑎 four, I’ve got the fourth term squared is equal to three hundred and twenty-four. Now taking the square roots of both sides, the square root of 𝑎 four squared is just 𝑎 four. And on the right hand-side, there’s two possible answers, a positive and a negative version of the square root three hundred and twenty-four, which is eighteen. So this means if my common ratio was positive, then the fourth term will be eighteen. But if the common ratio was negative, then my fourth term will be negative eighteen.

So let’s write that over here and then clear a bit of space for the next stage of our calculation. Now I think the easiest combination of numbers to use to work out my common ratio is the fifth term divided by the fourth term. So the fifth term is six, and the fourth term is either eighteen or negative eighteen. So my common ratio is either six over eighteen, or it’s six over negative eighteen. And they simplify to a third or negative a third. So I’ve got a couple of possibilities for the common ratio, and I know what the third term is. So I can use these to work out the possibilities for the second term and for the first term. And then, when I know that, I can work out my general formula.

So let’s start off by assuming that 𝑟 is equal to a third; the common ratio is a third. So if that’s the case, the third term is equal to the second time- second term times a third. And multiplying both sides by three would give me a second term of a hundred and sixty-two. And then following back one stage further, the second term is the first term times a third. So that’s a hundred and sixty-two is equal to the first term times a third. Now multiplying both sides of that by three would give me a first term of four hundred and eighty-six. So let’s just repeat that all now with 𝑟 is equal to negative a third. Then the third term will be the second term times negative a third. And that would mean that my second term is negative three times fifty-four, which is negative one hundred and sixty-two. So using this for my second term means that the first term is negative three times negative a hundred and sixty-two, which again is four hundred and eighty-six. So regardless of whether the common ratio was positive or negative, I’m still getting the first term in my sequence of four hundred and eighty-six.

So now I can work out the two possible formulae. In both cases, the first term is four hundred and eighty-six. And then I’ve got to do the version where the common ratio is a third, and the versh- version where the common ratio is negative a third. So using a common ratio of a third, the general term is the first term times 𝑟 to the power of 𝑛 minus one, so that the answer in this case will be four hundred and eighty-six times one third to the power of 𝑛 minus one.

So actually we got two possible answers, two possible general formulae from the information that we were given in the question. So we need to use both of those different formulae to work out the value of the eleventh term. So in each case, we’re gonna put 𝑛 equal to eleven. So in the first case, when the common ratio was positive a third, we end up with the eleventh term as four hundred and eighty-six times a third to the power of ten. And in the second case, where the ratio was negative a third, we end up with 𝑎 eleven. The eleventh term is equal to four hundred and eighty-six times negative a third to the power of ten. Well they’re both even powers. So whether the ratio is positive or negative, I’m gonna get the same answer cause all the negative signs are gonna cancel each other out. So no matter which formula we use, we’ve got an eleventh term of two over two hundred and forty-three.

Now let’s have a look at a couple of real-life situations. Harry invests 𝐼 dollars in an account that adds five percent interest every year. Find a formula telling us how much money will be in Harry’s account after 𝑛 years.

Now this is a geometric sequence, but we need to be really careful in this case with our starting conditions. What is the first term? Well, 𝐼 is the number of dollars invested at the very beginning at time zero. At the end of year one, we’ve added five percent. So a hundred percent plus five percent is a hundred and five percent. So to work out a hundred and five percent of something, we multiply it by one point o five. So Harry’s got 𝐼 times one point o five at the end of year one. So our first term 𝑎 one is 𝐼 times one point o five. Now at the end of the second year, we’ll have the amount that we had at the end of the first year, but we’ll have added another five percent to that. So we’ll times that by one point o five. And at the end of the third year, we’ll take our second year total and we’ll multiply that by one point o five. Now we can see a pattern emerging. At the end of the first year, we’ll have 𝐼 times one point o five to the power of one. At the end of the second year, we’ll have multiplied by one point o five twice, so we’ll have the initial amount times one point o five squared. And after the third year, we’ll have multiplied by one point o five three times, so the third amount will be 𝐼 times one point o five cubed. So the amount that we’ll have at the end of the 𝑛th year is 𝐼 times one point o five to the power of 𝑛. So that’s our formula.

Now the format of that general formula is slightly different to our normal geometric sequence formula. But if I adjusted the initial amount to be one point o five times the investment, because that’s the amount that we get at the end of year one, then I can put this more normal format here for our general term. So the thing is, when you’re dealing with these real-world situations, you just have to be really careful about the initial situation. So for example, if they asked us now how much was in the account after ten years, and they told us how much was initially invested, say eight thousand dollars, then we could use the formula like this. The amount at the end of year ten will be the initial amount, eight thousand, times one point o five, the multiplier for adding five percent of something, to the power of ten. So we’re doing that ten times. And since we are dealing with money, we should always round to two decimal places. So the amount that will be in there at the end of the ten years will be thirteen thousand and thirty-one dollars and sixteen cents.

So another real-life scenario then. An infectious disease is spreading through the population of rabbits on an island. The number of rabbits with the disease on a given day can be modelled by geometric sequence. At the end of day one, fifty rabbits have the disease. At the end of day eight, two thousand rabbits have the disease. How many will have the disease at the end of day nine?

Well let’s just do a little sketch. So we’ve got nine days. Day one, we’ve got fifty. Day eight, we’ve got two thousand. And each day, because this is a geometric sequence, we can multiply the number that have got the disease by some common ratio, let’s call it 𝑟, to work out how many will have the disease the next day. Well we can fill in the missing numbers. So the second day, we’ll have fifty times whatever that ratio is. The third day, we’ll have fifty times whatever that ratio is times whatever that ratio is. So fifty 𝑟 squared, and then fifty 𝑟 cubed, and-and so on. Now using that pattern, we know that fifty 𝑟 to the seven is two thousand. And dividing both sides by fifty, we know that 𝑟 to the power of seven is forty. So if I take the seventh root of both sides, I know that 𝑟 is equal to the seventh root of forty. Well I’m not gonna try and evaluate that at the moment, but I can use that here to work out how many rabbits will have the disease on the ninth day. It’s two thousand times that, which my calculator tells me is three thousand three hundred and eighty-seven point six two seven nine six and then lots of other digits. And in this context, I think it makes sense to round to the nearest whole number. So that’ll be three thousand three hundred and eighty-eight rabbits.