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Question Video: Identifying Geometric Sequences Mathematics • 9th Grade

Which of the following is π‘Ž geometric sequence? [A] π‘Ž_(𝑛) = 3(𝑛 + 3)Β², where 𝑛 β‰₯ 1. [B] π‘Ž_(𝑛) = 𝑛3^(𝑛 βˆ’ 1), where 𝑛 β‰₯ 2. [C] π‘Ž_(𝑛) = 𝑛(𝑛 + 2)Β², where 𝑛 β‰₯ 1. [D] π‘Ž_(𝑛) = 5π‘Ž_(𝑛 βˆ’ 1), where 𝑛 β‰₯ 2.

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Video Transcript

Which of the following is π‘Ž geometric sequence? Option (A) π‘Ž sub 𝑛 equals three times 𝑛 plus three squared, where 𝑛 is greater than or equal to one. Option (B) π‘Ž sub 𝑛 equals 𝑛 times three to the power of 𝑛 minus one, where 𝑛 is greater than or equal to two. Option (C) π‘Ž sub 𝑛 equals 𝑛 times 𝑛 plus two squared, where 𝑛 is greater than or equal to one. Option (D) π‘Ž sub 𝑛 equals five times π‘Ž sub 𝑛 minus one, where 𝑛 is greater than or equal to two.

We can begin by remembering that a geometric sequence is a sequence with a fixed ratio between successive or consecutive terms. So let’s take each of these sequences in the given answer options in turn. And it might be useful to find some of the first terms in each sequence in order to see if there is a fixed ratio between the terms.

So let’s begin with the sequence given in option (A), which has this 𝑛th term or π‘Ž sub 𝑛 as three times 𝑛 plus three squared. The first term in this sequence will be found when 𝑛 is equal to one. So that means that we can calculate the term with index one by substituting in a value of 𝑛 equals one. And so π‘Ž sub one must be three times one plus three squared. We can simplify one plus three as four. And then we’ll need to calculate three times four squared. Well, we know that four squared is 16. And when we multiply that by three, we get a value of 48. This means that this sequence will start with a value of 48.

So now, let’s work out the term which has an index of two. This time, we’ll be calculating three times two plus three squared. And this time, when we simplify two plus three, we get five, and we know that five squared is 25. So that gives us a second term in the sequence as 75. We can then work out one other term in this sequence; that is π‘Ž sub three. So when we calculate three times three plus three squared, that will give us a value of 108. So now, what we can do is look at these first three terms in the sequence and see if we can work out the ratio between the first term and the second term and the ratio between the second term and the third term.

If these two ratios are the same, then that would suggest that it is a geometric sequence. We can work out that if we multiplied the first term of 48 by the fraction 75 over 48, we would get the second term of 75. We can simplify this fraction to 25 over 16. Next, to go from the second term to the third term, we could multiply by 108 over 75. When we simplify this fraction, we get 36 over 25. We can now easily see that these two issues are not the same. Because they’re not the same, then there is not a fixed ratio between successive terms. And so option (A) is not a geometric sequence.

We can follow the same process to see if the sequence given in option (B) is geometric. This time, for the sequence, the index is given as 𝑛 is greater than or equal to two. This means that the term will begin with π‘Ž sub two. And so when we substitute 𝑛 is equal to two into the 𝑛th term, we get two times three to the power of two minus one. When we simplify this, we get a value of six. For the next term then, that’s the term which has an index of three, we’ll be calculating three times three to the power of three minus one. And when we simplify this, we get an answer of 27.

Next, the term with index four has a value of 108. We can then calculate the ratios between these terms. So the first ratio would be 27 over six. And that simplifies to nine over two. The next ratio between the terms of 108 and 27 would be simplified to four. If these two ratios were the same value, that would suggest it is a geometric sequence. However, as they are not, we can eliminate answer option (B).

For the 𝑛th term in option (C), we could observe that our index here is 𝑛 is greater than or equal to one. So let’s calculate the terms π‘Ž sub one, π‘Ž sub two, and π‘Ž sub three. When we substitute the values of 𝑛 equals one, two, and three into the 𝑛th term formula, we get values of nine, 32, and 75. We can once again look at the ratios between these three terms. We can work these out as 32 over nine and 75 over 32. Neither of these fractions simplifies any further. And we can see that they’re not equal. So option (C) is not a geometric sequence.

Finally, let’s have a look at the sequence which is given in option (D). This sequence is slightly different in that it’s an example of a recursive sequence. Because we have this 𝑛th term as π‘Ž sub 𝑛, then the term π‘Ž sub 𝑛 minus one is the term before π‘Ž sub 𝑛. In fact, if we wanted to describe this 𝑛th term in words, we could say that we find any term π‘Ž sub 𝑛 by multiplying the term before by five. If we wanted to find the first terms in this sequence, we may have a bit of a problem. That’s because even though we know that our index 𝑛 is greater than or equal to two, when we go to find π‘Ž sub two, we know that it must be five times π‘Ž sub one.

But we’re not actually told what π‘Ž sub one is. Usually, when we’re given the 𝑛th term of a sequence written in a recursive manner, then we also need to be told the value of the first term in the sequence or the term which has an index of one. But let’s see if it actually matters for this question.

We know that our sequence will have the terms π‘Ž sub two, π‘Ž sub three, and π‘Ž sub four. π‘Ž sub two will be five π‘Ž sub one, π‘Ž sub three will be equal to five π‘Ž sub two, and π‘Ž sub four will be five π‘Ž sub three. That means that the ratio between any successive term is five. And this means that we do have a sequence which has a fixed ratio between successive terms. Therefore, we can give the answer that the sequence π‘Ž sub 𝑛 equals five π‘Ž sub 𝑛 minus one where 𝑛 is greater than or equal to two is a geometric sequence.

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