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Question Video: Identifying a Sequence That Is Neither Arithmetic nor Geometric Mathematics • 9th Grade

Which of the following sequences is not classified as arithmetic or geometric? [A] 1/3, βˆ’1/3, βˆ’1, βˆ’5/3, ... [B] 1/2, 1/4, 1/8, 1/16, ... [C] 1/2, 1, 3/2, 2, ... [D] 1, 1/2, 0, βˆ’1/2, ... [E] 1/2, 1/3, 1/4, 1/5, ...

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Video Transcript

Which of the following sequences is not classified as arithmetic or geometric? Is it (A) one-third, negative one-third, negative one, negative five over three, and so on? (B) One-half, one-quarter, one-eighth, one sixteenth, and so on. (C) One-half, one, three over two, two, and so on. (D) One, one-half, zero, negative one-half, and so on. Or is it (E) one-half, one-third, one-quarter, one-fifth, and so on?

To start off with, we should probably remind ourselves what the terms arithmetic and geometric sequence mean. An arithmetic sequence is one in which each term can be obtained from a previous term by adding a common difference. In other words, if we subtract the 𝑛th term in the sequence from the 𝑛 plus oneth term, where 𝑛 is any natural number, the answer will always be a constant 𝑑. On the other hand, a geometric sequence is one where each term can be obtained from a previous term by multiplying by a common ratio. Mathematically speaking, if we divide the 𝑛 plus oneth term by the 𝑛th term for any natural number 𝑛, we will always get the constant π‘Ÿ.

With each of our options, we have been given four terms of the sequence, and the rest of the terms can be assumed to follow the same pattern. Therefore, the easiest way to classify the options is to verify in each case whether the terms we’ve been given satisfy either of the qualities shown on the right. Let us start by considering option (A).

First of all, to make things easier, we can label the terms as π‘Ž one, π‘Ž two, π‘Ž three, and π‘Ž four. We can then test whether the sequence is arithmetic by subtracting each term from the next term and seeing whether the difference is the same. For π‘Ž two minus π‘Ž one, we have negative one-third minus one-third, which is negative two-thirds. For π‘Ž three minus π‘Ž two, we have negative one minus negative one-third, which is the same thing as negative one plus a third, which is also negative two-thirds. Finally, just to be sure, we subtract π‘Ž four from π‘Ž three, which is negative five-thirds minus negative one, which is also negative two-thirds. We can see that the difference in each case is the same. So, we can conclude that the terms satisfy the requirements for being an arithmetic sequence. So, we can eliminate (A) as an option.

Next, let us consider (B). We can label the terms 𝑏 one, 𝑏 two, 𝑏 three, and 𝑏 four. In exactly the same way as before, we can consider the differences between successive terms to see whether the sequence is arithmetic. However, even if we just consider the first two differences, we can see that they are not the same. In fact, they are getting smaller. So, let us consider checking whether the sequence is geometric instead. We do this by taking successive ratios between the terms and comparing them, or 𝑏 two over 𝑏 one, we have one-quarter over one-half. We can evaluate this by taking the four into the denominator and two into the numerator, giving us two over four, which simplifies to a half.

Next, we have 𝑏 three over 𝑏 two, which is one-eighth over one-quarter. Using the same technique as before, this is four over eight, which is one-half again. We can then confirm that we also get one-half as the ratio for 𝑏 four over 𝑏 three. Since the ratio is the same between successive terms, this means we have a geometric sequence. So, let us also cross out (B) as an option.

Now that we are acquainted with the general procedure, let us consider an abbreviated approach for tackling the remaining options. For (C), we have the terms one-half, one, three over two, and two. We can see by inspection that each term can be obtained from the previous term by adding a half. Since there’s a common difference between terms, this means we have an arithmetic sequence. We could also verify this using the procedure from before by subtracting each term from the subsequent term in the sequence. So, option (C) is excluded.

Now, let us consider option (D). Listing out the terms one, a half, zero, and negative a half, we might notice that this is actually similar to the previous sequence. The difference is that this time we’re subtracting a half each time rather than adding a half. However, this is still an arithmetic sequence since the difference is still constant between successive terms. So, we can eliminate it.

At this point, we should expect our final option to be neither an arithmetic nor a geometric sequence, but let us check this for good measure. We will denote the terms as 𝑒 one to 𝑒 four to make things easier. Let us just take the first two pairs of terms for a common difference since we do not expect the sequence to be arithmetic. For 𝑒 two minus 𝑒 one, we have a third minus a half. We can subtract these fractions by giving them the same denominator. That is, we have two-sixths minus three-sixths, which is negative one-sixth. But for 𝑒 three minus 𝑒 two, we have a quarter minus a third, which is three twelfths minus four twelfths, which is negative one twelfth. We do not need to check the third difference, since it is already clear we do not have a constant difference between terms.

Let us now check whether the terms exhibit the geometric property. We first of all have 𝑒 two over 𝑒 one, which is a third over a half. By taking three into the denominator and two into the numerator, we get two-thirds. For the second ratio, we have one-quarter over one-third, and this is three-quarters. Since the ratio is not constant between terms, we cannot have a geometric sequence. In conclusion, since it is the only sequence that is neither arithmetic nor geometric, the answer is (E).

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