What is the common ratio of a geometric progression in which the second term is 6 and the 5th term is −162

To find the common ratio \( r \) of a geometric progression (GP), we can use the formula for the \( n \)-th term of a GP:

\[
a_n = a_1 \cdot r^{n-1}
\]

Given:
- The second term (\( a_2 \)) is 6, so \( a_2 = a_1 \cdot r = 6 \).
- The fifth term (\( a_5 \)) is \(-162\), so \( a_5 = a_1 \cdot r^4 = -162 \).

We have the following two equations:

\[
a_1 \cdot r = 6 \quad \text{(1)}
\]
\[
a_1 \cdot r^4 = -162 \quad \text{(2)}
\]

To find \( r \), divide equation (2) by equation (1):

\[
\frac{a_1 \cdot r^4}{a_1 \cdot r} = \frac{-162}{6}
\]

Simplifying:

\[
r^3 = \frac{-162}{6} = -27
\]

Now, take the cube root of both sides to find \( r \):

\[
r = \sqrt[3]{-27} = -3
\]

So, the common ratio \( r \) is \(-3\).

Would you like more details, or do you have any other questions?

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Here are 5 related questions:

1. How would you find the first term of this geometric progression?
2. What is the sum of the first 6 terms of this geometric progression?
3. How would the progression change if the common ratio was positive?
4. What is the general term (\( n \)-th term) formula for this specific progression?
5. Can you determine the 8th term of the progression?

**Tip:** In a geometric progression, the common ratio can be negative, which results in alternating positive and negative terms.

Learn how to find the common ratio of a geometric progression given the second term and the fifth term. This problem involves using the formula for the nth term of a geometric progression and solving equations. Suitable for students in Grades 9-12.

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