The geometric sequence is defined by the first term and the common ratio. Here, the first term \( a_1 \) is 1600 and the common ratio \( r \) is \( -\frac{1}{2} \). The \( n \)-th term of a geometric sequence can be expressed as:
\[ a_n = a_1 \cdot r^{n-1} \]
In this case, the \( n \)-th term is given by:
\[ a_n = 1600 \cdot \left(-\frac{1}{2}\right)^{n-1} \]
We need to find how many terms \( a_n \) are greater than 1000:
\[ |a_n| > 1000 \]
We start by calculating the absolute value of the \( n \)-th term, since the common ratio is negative, alternating the sign based on whether \( n-1 \) is even or odd:
\[ |a_n| = |1600 \cdot \left(-\frac{1}{2}\right)^{n-1}| = 1600 \cdot \left(\frac{1}{2}\right)^{n-1} \]
Next, we set up the inequality:
\[ 1600 \cdot \left(\frac{1}{2}\right)^{n-1} > 1000 \]
Dividing both sides by 1600 gives:
\[ \left(\frac{1}{2}\right)^{n-1} > \frac{1000}{1600} \] \[ \left(\frac{1}{2}\right)^{n-1} > \frac{5}{8} \]
Taking the logarithm of both sides (using any base, but we can use base 2 for convenience), we implement the change of base if necessary:
\[ n-1 < \log_{\frac{1}{2}}\left(\frac{5}{8}\right) \]
Using properties of logarithms, we can change the base:
\[ \log_{\frac{1}{2}}\left(x\right) = -\log_{2}\left(x\right) \]
Thus, we get:
\[ n-1 < -\log_{2}\left(\frac{5}{8}\right) = -\left(\log_{2}(5) - \log_{2}(8)\right) = -\log_{2}(5) + 3 \]
Now we can express \( n \):
\[ n < -\log_{2}(5) + 4 \]
Calculating \( \log_{2}(5) \):
\[ \log_{2}(5) \approx 2.32193 \]
So we compute:
\[ n < 4 - 2.32193 \approx 1.67807 \]
Since \( n \) must be a positive integer, the only integer satisfying this inequality is \( n = 1, 2, 3 \).
But we need to include negative terms based on the indexed nature of the common ratio. We check \( n = 1 \), \( 2 \), and \( 3 \):
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For \( n = 1 \): \[ a_1 = 1600 \] \( 1600 > 1000 \)
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For \( n = 2 \): \[ a_2 = 1600 \cdot \left(-\frac{1}{2}\right)^{1} = -800 \] \( -800 < 1000 \)
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For \( n = 3 \): \[ a_3 = 1600 \cdot \left(-\frac{1}{2}\right)^{2} = 400 \] \( 400 < 1000 \)
Continuing for later terms, we can see that the pattern of growth in absolute value continues to decrease as \( n \) increases through 400 terms.
The only viable terms that satisfy \( |a_n| > 1000 \) are \( a_1 = 1600 \). Hence, the only term greater than 1000 is the first term.
Thus, the final answer is:
\[ \boxed{1} \]