Example Sequence: 2, -4, 8, -16, 32, ... Explanation: See solution.
Practice makes perfect
We are asked to find a geometric sequence such that the second term is less than the first term, but the third one is greater than both of them.
a_2 < a_1 < a_3
Recall that every term in a geometric sequence is obtained by multiplying the previous one by the common ratio r.
Let's analyze how the terms in a geometric sequence behave for different values of the common ratio r.
Analyzing Geometric Sequences With Different Common Ratios
Let's start by considering the cases where the common ratio is positive.
One option is that r > 1. Then, each term is greater than the previous one. For example, r=2.
On the other hand, if 0 < r < 1, each term is smaller than the previous one. For example, r= 12.
As we can see, none of these options are helpful for our case. Therefore, r has to be negative. Once more there are two options. Let's consider that -1 < r < 0 first. For example, r=- 12.
Notice that in this case the starting term is the greatest in all the sequence and this is not what we want either. Finally, we can consider r < - 1. For example, r =-2.
In this sequence we can observe the desired behavior. Since the initial term is positive, the second term is less than the first, in this case - 4 < 2. Furthermore, the third term is greater than the second and the first terms, -4 < 2 < 8.
Conclusion
For a geometric sequence to have terms satisfying the given relationship,
a_2 < a_1 < a_3, the common ratio can be any number such that r < -1, provided that the first term is positive. Therefore there are infinitely many geometric sequences satisfying the given requirements. One example using r= - 2 is the sequence 2, -4, 8, -16, 32, ...
