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Question Video: Discussing the Monotonicity of a Geometric Sequence Mathematics • 9th Grade

For an increasing geometric sequence with first term π‘Ž and common ratio π‘Ÿ, which of the following could be true? [A] π‘Ž < βˆ’1, βˆ’1 < π‘Ÿ < 0 [B] π‘Ž > 0, 0 < π‘Ÿ < 1 [C] π‘Ž < 0, βˆ’1 < π‘Ÿ < 0 [D] π‘Ž < 0, 0 < π‘Ÿ < 1 [E] π‘Ž > 0, βˆ’1 < π‘Ÿ < 0

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Video Transcript

For an increasing geometric sequence with first term π‘Ž and common ratio π‘Ÿ, which of the following could be true? Is it (A) π‘Ž is less than negative one and π‘Ÿ is greater than negative one and less than zero? (B) π‘Ž is greater than zero and π‘Ÿ is between zero and one. Is it (C) π‘Ž is less than zero and π‘Ÿ is between negative one and zero? (D) π‘Ž is less than zero and π‘Ÿ is between zero and one. Or is it (E) π‘Ž is greater than zero and π‘Ÿ is between negative one and zero?

What do we mean when we talk about an increasing geometric sequence? A geometric sequence is one where each term is found by multiplying the previous term by a common ratio. The 𝑛th term of a geometric sequence is π‘Ž sub 𝑛 equals π‘Ž sub one times π‘Ÿ to the power of 𝑛 minus one, where π‘Ž sub one is the first term and π‘Ÿ is the common ratio. We say that this sequence is increasing if any term is greater than the previous, in other words, if π‘Ž sub 𝑛 plus one is greater than π‘Ž sub 𝑛 for all values of 𝑛. So, we’re going to use a bit of deduction here. We’re going to go through each example and establish whether that can be an increasing geometric sequence.

In our first sequence in option (A), we’re going to let π‘Ž sub one be equal to π‘Ž; that’s the first term. The common ratio is less than one; it’s between negative one and one. So, let’s think about that second term. It’s the product of the first term, which is π‘Ž, and the common ratio to the power of two minus one, which is just one. So, the second term is going to be π‘Žπ‘Ÿ. Now, since π‘Ž is less than negative one, it’s negative and π‘Ÿ is also negative, we can say that π‘Žπ‘Ÿ must be positive. And that’s really useful because it means that π‘Ž sub two must be greater than π‘Ž sub one.

But what happens when we try and find π‘Ž sub three? π‘Ž sub three is π‘Ž sub two times π‘Ÿ, so it’s π‘Ž times π‘Ÿ squared. Now, if we square π‘Ÿ, we end up with a positive value. So, we’re multiplying π‘Ž, which is negative, by a positive, meaning π‘Žπ‘Ÿ squared is going to be negative. Since π‘Ž sub two is positive and π‘Ž sub three is negative, it follows that π‘Ž sub three must be less than π‘Ž sub two. So, it does not follow that π‘Ž sub 𝑛 plus one is greater than π‘Ž sub one for all values of 𝑛. And so, we’re going to disregard option (A).

We’ll now consider (B). With option (B), π‘Ž is positive, as is the common ratio. Once again, the second term is π‘Žπ‘Ÿ. Now, this is a positive times a positive, so it’s also positive. But since π‘Ÿ is between zero and one, we can write it as some fraction one over π‘₯ for values of π‘₯ greater than one, meaning that π‘Ž sub two is equal to π‘Ž over π‘₯, which by definition absolutely has to be less than π‘Ž. So, π‘Ž sub two must be less than π‘Ž sub one. And we’ve shown straightaway that our function cannot be increasing, so we disregard option (B).

So, what about option (C)? Once again, our first term is π‘Ž and our second term is π‘Žπ‘Ÿ. This time π‘Ž is once again negative as is the common ratio, which means π‘Ž times π‘Ÿ is positive. However, just in our first example, if we then multiply this by a common ratio, which is negative, we end up with a negative value. And so, π‘Ž sub three is less than π‘Ž sub two. And our sequence in option (C) cannot be increasing.

So option (D), we start with the same two pieces of working: π‘Ž sub two is π‘Ž times the common ratio. Now, since π‘Ž is less than zero, multiplying it by a ratio between zero and one is also less than zero. That’s not enough to help us decide whether π‘Ž sub two is greater than or smaller than π‘Ž sub one. But we do know that the common ratio is between zero and one. So once again, we’ll rewrite our common ratio as one over π‘₯ for values of π‘₯ greater than one, and π‘Ž sub two is equal to π‘Ž over π‘₯. Now, since π‘Ž is negative, π‘Ž over π‘₯ while smaller in magnitude is actually greater than the value of π‘Ž. So, it follows that π‘Ž sub two is greater than π‘Ž sub one. In a similar way, π‘Ž sub three is π‘Ž over π‘₯ squared. Since π‘Ž is negative and π‘₯ is greater than one whilst π‘Ž over π‘₯ squared is smaller in magnitude than π‘Ž over π‘₯, it’s actually sits above it on the number line. So once again, π‘Ž sub three is greater than π‘Ž sub two.

We can carry on in this manner. Essentially, each time we multiply by one over π‘₯, whilst the magnitude of the number gets smaller because π‘Ž is negative and π‘₯ is greater than one, π‘Ž over π‘₯ to the 𝑛th power will always be greater than π‘Ž over π‘₯ to the 𝑛 minus oneth power. So, for any value of 𝑛, π‘Ž sub 𝑛 plus one will always be greater than π‘Ž sub 𝑛. This means that the answer is (D). An increasing geometric sequence with first term π‘Ž and common ratio π‘Ÿ is π‘Ž is less than zero and π‘Ÿ is between zero and one.

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