The problem is asking to find the common ratio ($r$) of a geometric sequence where the second term ($a_2$) is $-6$ and the fifth term ($a_5$) is $48$.

In a geometric sequence, the $n$-th term is given by the formula: $a_n = a_1 \cdot r^{n-1}$ where:

• $a_n$ is the $n$-th term,
• $a_1$ is the first term,
• $r$ is the common ratio, and
• $n$ is the position of the term.

We are given:

• $a_2 = -6$,
• $a_5 = 48$.

We can set up two equations:

1. $a_2 = a_1 \cdot r^1 = -6$,
2. $a_5 = a_1 \cdot r^4 = 48$.

Step 1: Solve for $a_1$ and $r$

From the first equation, $a_1 \cdot r = -6$, so: $a_1 = \frac{-6}{r}$

Substitute this into the second equation: $\left(\frac{-6}{r}\right) \cdot r^4 = 48$ Simplify: $-6 \cdot r^3 = 48$ $r^3 = \frac{48}{-6} = -8$

Step 2: Solve for $r$

$r = \sqrt[3]{-8} = -2$

Thus, the common ratio of the geometric sequence is $r = -2$.

Would you like more details on any step? Here are some related questions:

1. How do you derive the formula for the nth term of a geometric sequence?
2. What are the conditions for a geometric sequence to be convergent or divergent?
3. How would the process change if we had different terms given?
4. Can you explain why a negative ratio causes alternating terms?
5. How does the formula for geometric sequences apply to infinite series?

Tip: Always verify your result by substituting back into the original equations to ensure accuracy.