# Question Video: Finding the First Terms of a Geometric Sequence Mathematics

In a geometric sequence, and for a certain π, we have π_(π) = 6, π_(π + 1) = 3, and π_(2π β 1) = 3/8. Find the first 3 terms of the sequence and what π is.

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### Video Transcript

In a geometric sequence, and for a certain π, we have π sub π equals six, π sub π plus one equals three, and π sub two π minus one equals three-eighths. Find the first three terms of the sequence and what π is.

In this question, weβre told that we have a geometric sequence, and thatβs a sequence where there is a fixed ratio between successive terms. Letβs recap a little bit of the notation around geometric sequences, beginning with the fact that we can note each term in terms of its index. For example, the first term can be written as π or π sub one, the second term as π sub two, and the third term as π sub three. And in a geometric sequence because we multiply any term by π to find the next term, then we could also write that the terms could be written as π then ππ then ππ squared, and so on. This leads us into the πth term formula for geometric sequence: π sub π is equal to π times π to the power of π minus one, where π is the index, π is the first term, and π is the fixed ratio between terms.

So, letβs have a look at what weβre told about the terms of this sequence. The first piece of information weβre given is that for a certain value of π we have π sub π is equal to six. So, somewhere in this sequence there is a term six. Weβre then given another term, π sub π plus one is equal to three, so we know that somewhere in this sequence there will be a term three. However, π sub π plus one is the term that comes immediately after π sub π. And so, we know that three must come immediately after the term six. Weβre then given a third term. π sub two π minus one is equal to three eighths. We donβt know exactly where this fits in relation to the two other given terms, so letβs just note that itβs somewhere else in the sequence.

Now, letβs see if we can write these three terms in a way that relates π and π. We can write the first of our given terms π sub π as π times π to the power of π minus one. The next term π sub π plus one can be written as π times π to the power of π. Remember that this comes from the fact that the exponent of π must be one less than the index. The final given term π sub two π minus one can be written as π times π to the power of two π minus two.

At the minute, we have a lot of unknowns in this question. However, we can work out the value of π given that we have two successive terms in the sequence. In a geometric sequence, because we could take any term and multiply it by the fixed ratio to get the next term, then we can ask ourselves, what must we multiply six by to get three? Well, it must be one-half. And so, we have our first piece of information that π is equal to one-half. Since we know that π is equal to one-half, we can rewrite π times π to the power of π minus one as π times one-half to the power of π minus one, and we know that it must equal six.

Then using the fact again that π is equal to one-half, we can write a second equation that π times one-half to the power of π is equal to three. We could also write a third equation that π times one-half to the power of two π minus two is equal to three-eighths. At the minute, we couldnβt use any single one of these equations to solve since we have two unknowns π and π. We can however solve two of these equations simultaneously. So, letβs see if we could take any of the equations and write it in terms of π.

Using the second equation, we could divide through by one-half to the power of π. We can write the denominator of this fraction in a nicer way by recognizing that if weβre dividing by a number to the power of π, then thatβs equivalent to multiplying by that number to the power of negative π. Now, we have a value for π in terms of π. And letβs substitute this into our third equation. This gives us three times one-half to the power of negative π times one-half to the power of two π minus two is equal to three-eighths. In order to simplify this, we can remember one of the rules of exponents that π to the power of π₯ multiplied by π to the power of π¦ is equal to π to the power of π₯ plus π¦. And therefore, when we add the exponents negative π and two π minus two, we get an exponent of π minus two. We can then divide both sides by three. And three-eights divided by three is three twenty-fourths. We can further simplify this to give us one-half to the power of π minus two is equal to one-eighth.

At this point, there are a number of different ways we could potentially solve this to find the value of π. However, perhaps the easiest way is simply to consider the whole exponent as an unknown value. And therefore, weβre asking one-half to the power of what would give us one-eighth. And in fact, one-half cubed would give us one-eighth since one-half times one-half is one-quarter and one-quarter times one-half is one-eighth. And therefore, we can say that this exponent must be equal to three. So, π minus two is equal to three. And so, π must be equal to five.

So now weβve worked out the value of π, letβs clear some space and see how this has helped us working out the sequence. Given that π is equal to five means that we can say that itβs the fifth term, which is equal to six. We can then substitute π is equal to five into this equation, π times one-half to the power of π minus one is equal to six. We can then simplify the exponent. When we calculate one-half to the power of four, we get one over 16. We can then divide both sides by one over 16, which is equivalent to multiplying by 16. And six times 16 is 96. So now, we know the value π is equal to 96.

Weβve now found the value of all of the unknowns in this question. However, the fact that π is equal to 96 also gives us the first term of the sequence. The first term of this sequence must be 96. Weβre asked here to find the first three terms, so we need to find the second and third term. The second term in the sequence will be given by the first term multiplied by the common ratio π. And since 96 multiplied by one-half is 48, then the second term is 48. We can work out the third term of this sequence in two different ways. The first way is by recognizing that it must be equal to 96 multiplied by one-half squared. Or alternatively, we could also realize that to find the third term we take the second term and multiply it by the common ratio of one-half. Either way would give us the value of 24.

We can then give a complete answer. We were asked for the first three terms of the sequence, so thatβs 96, 48, 24. We were then asked to find the value of π, and so π is equal to five.