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.

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.