Question
How many terms of the sequence 3, 9, 27, ... are needed to exceed a sum of 1000?
Answers
GPT-4o mini
The given sequence is a geometric sequence where the first term \( a = 3 \) and the common ratio \( r = 3 \). The \( n \)-th term of the sequence can be expressed as:
\[
a_n = 3 \cdot 3^{n-1} = 3^n
\]
We need to find the smallest \( n \) such that the sum of the first \( n \) terms exceeds 1000. The sum \( S_n \) of the first \( n \) terms of a geometric sequence can be calculated using the formula:
\[
S_n = a \frac{r^n - 1}{r - 1}
\]
In our case:
\[
S_n = 3 \frac{3^n - 1}{3 - 1} = 3 \frac{3^n - 1}{2} = \frac{3^{n+1} - 3}{2}
\]
We want to solve the inequality:
\[
\frac{3^{n+1} - 3}{2} > 1000
\]
To eliminate the fraction, we multiply both sides by 2:
\[
3^{n+1} - 3 > 2000
\]
Next, adding 3 to both sides gives:
\[
3^{n+1} > 2003
\]
Now, we need to find \( n \) such that this holds true. We can take the logarithm of both sides:
\[
\log(3^{n+1}) > \log(2003)
\]
Using the property of logarithms, this becomes:
\[
(n+1) \log(3) > \log(2003)
\]
Solving for \( n+1 \):
\[
n + 1 > \frac{\log(2003)}{\log(3)}
\]
To find the value of \( n \), we need to compute \( \log(2003) \) and \( \log(3) \). Using approximate values,
\[
\log(3) \approx 0.4771 \quad \text{and} \quad \log(2003) \approx 3.301
\]
Calculating \( \frac{\log(2003)}{\log(3)} \):
\[
\frac{3.301}{0.4771} \approx 6.91
\]
Thus,
\[
n + 1 > 6.91 \quad \Rightarrow \quad n > 5.91
\]
Since \( n \) must be a whole number, we take \( n = 6 \). Let's verify if \( n = 6 \) is the correct choice.
Now we calculate \( S_6 \):
\[
S_6 = \frac{3^{6 + 1} - 3}{2} = \frac{3^7 - 3}{2} = \frac{2187 - 3}{2} = \frac{2184}{2} = 1092
\]
Since \( S_6 = 1092 > 1000 \), we check \( S_5 \):
\[
S_5 = \frac{3^{5 + 1} - 3}{2} = \frac{3^6 - 3}{2} = \frac{729 - 3}{2} = \frac{726}{2} = 363
\]
Since \( S_5 = 363 < 1000 \), we conclude that the answer is:
\[
\boxed{6}
\]
Thus, 6 terms of the sequence are needed to exceed a sum of 1000.
\[
a_n = 3 \cdot 3^{n-1} = 3^n
\]
We need to find the smallest \( n \) such that the sum of the first \( n \) terms exceeds 1000. The sum \( S_n \) of the first \( n \) terms of a geometric sequence can be calculated using the formula:
\[
S_n = a \frac{r^n - 1}{r - 1}
\]
In our case:
\[
S_n = 3 \frac{3^n - 1}{3 - 1} = 3 \frac{3^n - 1}{2} = \frac{3^{n+1} - 3}{2}
\]
We want to solve the inequality:
\[
\frac{3^{n+1} - 3}{2} > 1000
\]
To eliminate the fraction, we multiply both sides by 2:
\[
3^{n+1} - 3 > 2000
\]
Next, adding 3 to both sides gives:
\[
3^{n+1} > 2003
\]
Now, we need to find \( n \) such that this holds true. We can take the logarithm of both sides:
\[
\log(3^{n+1}) > \log(2003)
\]
Using the property of logarithms, this becomes:
\[
(n+1) \log(3) > \log(2003)
\]
Solving for \( n+1 \):
\[
n + 1 > \frac{\log(2003)}{\log(3)}
\]
To find the value of \( n \), we need to compute \( \log(2003) \) and \( \log(3) \). Using approximate values,
\[
\log(3) \approx 0.4771 \quad \text{and} \quad \log(2003) \approx 3.301
\]
Calculating \( \frac{\log(2003)}{\log(3)} \):
\[
\frac{3.301}{0.4771} \approx 6.91
\]
Thus,
\[
n + 1 > 6.91 \quad \Rightarrow \quad n > 5.91
\]
Since \( n \) must be a whole number, we take \( n = 6 \). Let's verify if \( n = 6 \) is the correct choice.
Now we calculate \( S_6 \):
\[
S_6 = \frac{3^{6 + 1} - 3}{2} = \frac{3^7 - 3}{2} = \frac{2187 - 3}{2} = \frac{2184}{2} = 1092
\]
Since \( S_6 = 1092 > 1000 \), we check \( S_5 \):
\[
S_5 = \frac{3^{5 + 1} - 3}{2} = \frac{3^6 - 3}{2} = \frac{729 - 3}{2} = \frac{726}{2} = 363
\]
Since \( S_5 = 363 < 1000 \), we conclude that the answer is:
\[
\boxed{6}
\]
Thus, 6 terms of the sequence are needed to exceed a sum of 1000.