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

For an increasing geometric sequence with first term π and common ratio π, which of the following could be true? [A] π < β1, β1 < π < 0 [B] π > 0, 0 < π < 1 [C] π < 0, β1 < π < 0 [D] π < 0, 0 < π < 1 [E] π > 0, β1 < π < 0

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### Video Transcript

For an increasing geometric sequence with first term π and common ratio π, which of the following could be true? Is it (A) π is less than negative one and π is greater than negative one and less than zero? (B) π is greater than zero and π is between zero and one. Is it (C) π is less than zero and π is between negative one and zero? (D) π is less than zero and π is between zero and one. Or is it (E) π is greater than zero and π 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 πth term of a geometric sequence is π sub π equals π sub one times π to the power of π minus one, where π sub one is the first term and π is the common ratio. We say that this sequence is increasing if any term is greater than the previous, in other words, if π sub π plus one is greater than π sub π for all values of π. So, weβre going to use a bit of deduction here. Weβre going to go through each example and establish whether that can be an increasing geometric sequence.

In our first sequence in option (A), weβre going to let π sub one be equal to π; thatβs the first term. The common ratio is less than one; itβs between negative one and one. So, letβs think about that second term. Itβs the product of the first term, which is π, and the common ratio to the power of two minus one, which is just one. So, the second term is going to be ππ. Now, since π is less than negative one, itβs negative and π is also negative, we can say that ππ must be positive. And thatβs really useful because it means that π sub two must be greater than π sub one.

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

Weβll now consider (B). With option (B), π is positive, as is the common ratio. Once again, the second term is ππ. Now, this is a positive times a positive, so itβs also positive. But since π 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 sub two is equal to π over π₯, which by definition absolutely has to be less than π. So, π sub two must be less than π sub one. And weβ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 and our second term is ππ. This time π is once again negative as is the common ratio, which means π times π 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 sub three is less than π sub two. And our sequence in option (C) cannot be increasing.

So option (D), we start with the same two pieces of working: π sub two is π times the common ratio. Now, since π is less than zero, multiplying it by a ratio between zero and one is also less than zero. Thatβs not enough to help us decide whether π sub two is greater than or smaller than π sub one. But we do know that the common ratio is between zero and one. So once again, weβll rewrite our common ratio as one over π₯ for values of π₯ greater than one, and π sub two is equal to π over π₯. Now, since π is negative, π over π₯ while smaller in magnitude is actually greater than the value of π. So, it follows that π sub two is greater than π sub one. In a similar way, π sub three is π over π₯ squared. Since π is negative and π₯ is greater than one whilst π over π₯ squared is smaller in magnitude than π over π₯, itβs actually sits above it on the number line. So once again, π sub three is greater than π 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 is negative and π₯ is greater than one, π over π₯ to the πth power will always be greater than π over π₯ to the π minus oneth power. So, for any value of π, π sub π plus one will always be greater than π sub π. This means that the answer is (D). An increasing geometric sequence with first term π and common ratio π is π is less than zero and π is between zero and one.