### 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 πth 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, weβre going to use a bit of deduction here. Weβre going to go through each example and establish whether that can be an increasing geometric sequence.

In our first sequence in option (A), weβre going to let π sub one be equal to π; thatβs the first term. The common ratio is less than one; itβs between negative one and one. So, letβs think about that second term. Itβs 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 ππ. Now, since π is less than negative one, itβs negative and π is also negative, we can say that ππ must be positive. And thatβs 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 itβs π times π squared. Now, if we square π, we end up with a positive value. So, weβre multiplying π, which is negative, by a positive, meaning ππ 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, weβre going to disregard option (A).

Weβll now consider (B). With option (B), π is positive, as is the common ratio. Once again, the second term is ππ. Now, this is a positive times a positive, so itβs 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 weβve 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 ππ. 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. Thatβs 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, weβll 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₯, itβs 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 πth 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.