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Question Video: Identifying Geometric Sequences Mathematics • 9th Grade

Determine 𝑝 and π‘ž that make the following a geometric sequence, starting at π‘Žβ‚, of the form π‘Ž_(𝑛) = π‘π‘ž^(𝑛 βˆ’ 1): 2, 6, 18, 54, 162, ...

02:34

Video Transcript

Determine 𝑝 and π‘ž that make the following a geometric sequence, starting at π‘Ž sub one, of the form π‘Ž sub 𝑛 equals 𝑝 times π‘ž to the power of 𝑛 minus one: two, six, 18, 54, 162, and so on.

Let’s start by recalling that a geometric sequence is a sequence which has a fixed ratio between successive terms. In other words, if you perform the same multiplication on any term, you will get the next term. So let’s consider how we would go from two to six. Well, we could do this by multiplying by three. Notice that if we were adding or subtracting values to two, then this wouldn’t be a geometric sequence. It would in fact be an arithmetic sequence. So if we also multiplied six by three, would we get 18? Yes, we would. In the same way, if we multiply 18 by three, we get 54. And 54 multiplied by three does give us 162. So it looks as though we do have a geometric sequence.

So the next thing to do is see if we can write it in this form of the 𝑛th term, which involves 𝑝 and π‘ž. We should remember that we can write any term π‘Ž sub 𝑛 in a geometric sequence as π‘Ž times π‘Ÿ to the power of 𝑛 minus one, where 𝑛 is the index, π‘Ž or π‘Ž sub one represents the first term, and π‘Ÿ is the fixed ratio between terms. It’s worth noting that sometimes we have sequences that begin with index zero. However, here, we’re given that this sequence should start with an index of one with the term π‘Ž sub one.

So let’s compare this standard way of writing the 𝑛th term with the form in which we were given, which involves 𝑝 and π‘ž. When we look at the equivalent values in the general 𝑛th term formula, we could see that 𝑝 must be equivalent to the first term. π‘ž would be equivalent to the common ratio or fixed ratio between terms. So let’s consider what the value of 𝑝 would be for this given sequence. It’s quite easy to spot. It’s the first term. And so 𝑝 must be equal to two. We then determined that π‘ž must be equivalent to the fixed ratio between terms. And we’ve already worked out that the fixed ratio must be three.

We can therefore give the answer that 𝑝 is equal to two and π‘ž is equal to three.

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