In a geometric sequence, the general formula is given by:
\[ a_n = a_1 \cdot r^{n-1} \]
where:
- \( a_n \) is the \( n \)-th term,
- \( a_1 \) is the first term,
- \( r \) is the common ratio, and
- \( n \) is the term number.
From the problem, we know:
- \( a_2 = -144 \)
- \( a_5 = 486 \)
Using the general formula, we can express these terms as follows:
-
For \( a_2 \): \[ a_2 = a_1 \cdot r^{1} = -144 \quad \text{(1)} \]
-
For \( a_5 \): \[ a_5 = a_1 \cdot r^{4} = 486 \quad \text{(2)} \]
Now, we can solve these equations to find \( a_1 \) and \( r \).
From equation (1): \[ a_1 \cdot r = -144 \quad \Rightarrow \quad a_1 = \frac{-144}{r} \]
Substituting \( a_1 \) into equation (2): \[ \frac{-144}{r} \cdot r^{4} = 486 \] \[ -144 r^{3} = 486 \] \[ r^{3} = \frac{486}{-144} = -\frac{27}{8} \]
Now, we find \( r \) by taking the cube root: \[ r = -\left(\frac{27}{8}\right)^{1/3} = -\frac{3}{2} \]
Now substitute \( r = -\frac{3}{2} \) back into equation (1) to find \( a_1 \): \[ a_1 \cdot \left(-\frac{3}{2}\right) = -144 \] \[ a_1 = -144 \cdot \left(-\frac{2}{3}\right) = 96 \]
Thus, the values we found are:
- \( a_1 = 96 \)
- \( r = -\frac{3}{2} \)
Therefore, the explicit formula for the geometric sequence is: \[ a_n = 96 \left(-\frac{3}{2}\right)^{n-1} \]
So the correct answer is: \[ \boxed{a_n = 96 \left(-\frac{3}{2}\right)^{n-1}} \]
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