Sure! This problem describes a geometric sequence, where each term is found by multiplying the previous term by the common ratio.

Given:

• First term $a_1 = 15$
• Common ratio $r = -3$

The formula for the nth term of a geometric sequence is: $a_n = a_1 \cdot r^{n-1}$

Now, let’s find the first five terms:

1. First term $a_1 = 15$
2. Second term $a_2 = 15 \times (-3) = -45$
3. Third term $a_3 = -45 \times (-3) = 135$
4. Fourth term $a_4 = 135 \times (-3) = -405$
5. Fifth term $a_5 = -405 \times (-3) = 1215$

The first five terms are:

$15, -45, 135, -405, 1215$

If you want to write it neatly on paper, you could lay it out like this:

$a_1 = 15, \quad a_2 = -45, \quad a_3 = 135, \quad a_4 = -405, \quad a_5 = 1215$

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Would you like a more detailed explanation or clarification? Also, here are some related questions for further practice:

1. How do you calculate the nth term of a geometric sequence?
2. What happens if the common ratio is a fraction or a decimal?
3. How can you identify the terms in a geometric sequence if only the common ratio and a few terms are given?
4. How would you write the general formula for a geometric sequence?
5. How do geometric sequences relate to exponential functions?

Tip: Always double-check your multiplication when working with negative common ratios to ensure you don't make sign errors!