Question

A geometric sequence has 400 terms. The first term is 1600 and the common ratio is $-\frac{1}{2}.$ How many terms of this sequence are greater than 1000?

Answer 1

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 \( 3 \):

1. For \( n = 1 \):
\[
a_1 = 1600
\]
\( 1600 > 1000 \)

2. For \( n = 2 \):
\[
a_2 = 1600 \cdot \left(-\frac{1}{2}\right)^{1} = -800
\]
\( -800 < 1000 \)

3. 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:

\[
\boxed{1}
\]