The given sequence is $a_n = (1.04)^{n-1}$, which represents a geometric sequence. In a geometric sequence, the common ratio is the constant factor between consecutive terms, and the first term is obtained when $n = 1$.

Solution:

1. Common Ratio ($r$): In any geometric sequence given in the form $a_n = ar^{n-1}$, $r$ is the base of the exponential function. Here, $r = 1.04$.

2. First Four Terms ($a_1, a_2, a_3, a_4$):

   • $a_1 = (1.04)^{1-1} = (1.04)^0 = 1$
   • $a_2 = (1.04)^{2-1} = (1.04)^1 = 1.04$
   • $a_3 = (1.04)^{3-1} = (1.04)^2 = 1.0816$
   • $a_4 = (1.04)^{4-1} = (1.04)^3 = 1.124864$

Final Answer:

• Common Ratio: $r = 1.04$
• $a_1 = 1$
• $a_2 = 1.04$
• $a_3 = 1.0816$
• $a_4 = 1.124864$

Would you like detailed explanations for how these terms were calculated, or have any other questions?

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Related Questions:

1. How can you generalize a geometric sequence formula for any term $n$?
2. What happens to the terms of this sequence as $n \to \infty$?
3. What is the sum of the first 10 terms of this sequence?
4. How would you express the $n$-th term if the common ratio was negative, e.g., $r = -1.04$?
5. How do geometric sequences differ from arithmetic sequences?

Tip: In geometric sequences, multiplying by the common ratio repeatedly builds the sequence—knowing $r$ allows you to predict any term!