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We are given a recursive sequence where:

\[
a_1 = 2
\]
\[
a_n = -9a_{n-1}
\]

We need to find \( a_2 \), \( a_3 \), and \( a_4 \).

### Step 1: Find \( a_2 \)
Using the recursive formula \( a_n = -9a_{n-1} \), we substitute \( n = 2 \):

\[
a_2 = -9a_1 = -9(2) = -18
\]

### Step 2: Find \( a_3 \)
Using the same formula for \( n = 3 \):

\[
a_3 = -9a_2 = -9(-18) = 162
\]

### Step 3: Find \( a_4 \)
Now for \( n = 4 \):

\[
a_4 = -9a_3 = -9(162) = -1458
\]

### Final Answers:
\[
a_2 = -18
\]
\[
a_3 = 162
\]
\[
a_4 = -1458
\]

Let me know if you want further details on these calculations.

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Here are five related questions:
1. How can the recursive formula be generalized for any \( n \)?
2. What is the common ratio between consecutive terms in this sequence?
3. How would the sequence change if \( a_1 \) had a different value?
4. Can you express a closed-form solution for this sequence?
5. How does this sequence behave as \( n \to \infty \)?

**Tip**: Recursive sequences rely heavily on prior terms. Knowing the base case is crucial for solving them.

Find a_2, a_3, and a_4. Given: a_1 = 2, a_n = -9a_(n-1).

Learn how to solve a recursive sequence where a_1 = 2 and a_n = -9a_(n-1). The solution finds a_2, a_3, and a_4 using the given recursive formula. Ideal for students in Grades 9-11.

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