In a geometric series, the $n$-th term $T_n$ is given by the formula:

$T_n = a \cdot r^{n-1}$

where:

• $a$ is the first term,
• $r$ is the common ratio,
• $n$ is the term number.

Given:

• The 5th term $T_5 = -2$,
• The 9th term $T_9 = 8\sqrt{2}$.

From the general formula, we have:

1. For the 5th term: $T_5 = a \cdot r^{5-1} = a \cdot r^4 = -2$ So, we get the equation: a \cdot r^4 = -2 \tag{1}

2. For the 9th term: $T_9 = a \cdot r^{9-1} = a \cdot r^8 = 8\sqrt{2}$ So, we get the equation: a \cdot r^8 = 8\sqrt{2} \tag{2}

Now, we have the system of two equations:

1. $a \cdot r^4 = -2$
2. $a \cdot r^8 = 8\sqrt{2}$

Solving for the ratio $r$:

To eliminate $a$, divide equation (2) by equation (1):

$\frac{a \cdot r^8}{a \cdot r^4} = \frac{8\sqrt{2}}{-2}$

This simplifies to:

$r^4 = \frac{8\sqrt{2}}{-2} = -4\sqrt{2}$

So,

r^4 = -4\sqrt{2} \tag{3}

Taking the fourth root of both sides:

$r = \sqrt[4]{-4\sqrt{2}}$

Since $r^4$ is a negative number, $r$ must be a complex number. Let's simplify the fourth root:

[ r = \sqrt[4]{-4\sqrt{2}} = \pm \left( \sqrt[4]{4} \cdot \sqrt[4]{\sqrt{2}} \right) ] $r = \pm \left( \sqrt{2} \cdot 2^{1/8} \right)$

So the ratio $r$ will have a real and imaginary part in this form.

Let me know if you need further elaboration or have any follow-up questions.

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Related Questions:

1. How do you find the 6th term in this geometric series?
2. What happens if $r$ is a real number in a geometric sequence?
3. How can we use complex numbers to solve geometric series problems?
4. What does the convergence of a geometric series look like when $|r| < 1$?
5. How do you find the sum of an infinite geometric series?

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Tip:

When dealing with geometric series problems, always try to express terms in terms of powers of the common ratio, especially when solving for unknowns like the first term or ratio. This will help simplify equations and allow for easier manipulation.