In a geometric progression, the terms are related by a constant ratio, denoted as $r$. Let's denote the first term of the sequence as $a$.

The terms of a geometric progression can be written as:

• First term: $a$
• Second term: $a \cdot r$
• Third term: $a \cdot r^2$
• Fourth term: $a \cdot r^3$

We are given that:

1. The second term is 4, so $a \cdot r = 4$
2. The fourth term is 9, so $a \cdot r^3 = 9$

We can now set up the following system of equations:

$a \cdot r = 4 \quad \text{(1)}$ $a \cdot r^3 = 9 \quad \text{(2)}$

To solve for $r$, divide equation (2) by equation (1):

$\frac{a \cdot r^3}{a \cdot r} = \frac{9}{4}$ This simplifies to: $r^2 = \frac{9}{4}$ Taking the square root of both sides: $r = \pm \frac{3}{2}$

Now, substitute this value of $r$ back into equation (1) to solve for $a$:

If $r = \frac{3}{2}$: $a \cdot \frac{3}{2} = 4 \quad \Rightarrow \quad a = \frac{4 \cdot 2}{3} = \frac{8}{3}$

If $r = -\frac{3}{2}$: $a \cdot \left(-\frac{3}{2}\right) = 4 \quad \Rightarrow \quad a = \frac{4 \cdot 2}{-3} = \frac{-8}{3}$

Now, calculate the third term $a \cdot r^2$:

1. For $r = \frac{3}{2}$: $\text{Third term} = a \cdot r^2 = \frac{8}{3} \cdot \frac{9}{4} = 6$

2. For $r = -\frac{3}{2}$: $\text{Third term} = a \cdot r^2 = \frac{-8}{3} \cdot \frac{9}{4} = -6$

Thus, the third term is either $6$ or $-6$, depending on the sign of the common ratio.

Would you like more details on how to interpret these results?

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Here are 5 related questions to explore:

1. How do you find the sum of a geometric progression?
2. What happens when the common ratio $r$ is a negative value?
3. How do you determine the number of terms in a geometric progression?
4. How do you solve geometric progression problems when given non-consecutive terms?
5. Can a geometric progression have a common ratio greater than 1?

Tip: Always check if the common ratio is positive or negative in geometric progression problems, as it can affect the signs of the terms.