Question
Wanahton purified a portion of water with
\[900\] grams of contaminants. Each hour, a third of the contaminants was filtered out.
Let
\[g(n)\] be the amount of contaminants (in grams) that remained by the beginning of the
\[n^\text{th}\] hour.
\[g\] is a sequence. What kind of sequence is it?
Choose 1 answer:
Choose 1 answer:
(Choice A) Arithmetic sequence
A
Arithmetic sequence
(Choice B) Geometric sequence
B
Geometric sequence
Write an explicit formula for the sequence.
\[g(n)=\]
Related content
\[900\] grams of contaminants. Each hour, a third of the contaminants was filtered out.
Let
\[g(n)\] be the amount of contaminants (in grams) that remained by the beginning of the
\[n^\text{th}\] hour.
\[g\] is a sequence. What kind of sequence is it?
Choose 1 answer:
Choose 1 answer:
(Choice A) Arithmetic sequence
A
Arithmetic sequence
(Choice B) Geometric sequence
B
Geometric sequence
Write an explicit formula for the sequence.
\[g(n)=\]
Related content
Answers
GPT-4o mini
To determine the type of sequence represented by \( g(n) \), we analyze how the contaminants are filtered out.
Initially, the amount of contaminants is \( g(0) = 900 \) grams. Each hour, one-third of the contaminants is filtered out. This implies that two-thirds of the contaminants remain each hour.
Mathematically, we can express the amount of contaminants remaining at the beginning of the \( n^\text{th} \) hour as follows:
1. After the first hour, the amount remaining is:
\[
g(1) = \frac{2}{3} \cdot g(0) = \frac{2}{3} \cdot 900 = 600 \text{ grams}
\]
2. After the second hour:
\[
g(2) = \frac{2}{3} \cdot g(1) = \frac{2}{3} \cdot 600 = 400 \text{ grams}
\]
3. After the third hour:
\[
g(3) = \frac{2}{3} \cdot g(2) = \frac{2}{3} \cdot 400 \approx 266.67 \text{ grams}
\]
From this, we see that each term is produced by multiplying the previous term by a constant factor, \(\frac{2}{3}\). Thus, this sequence is a **geometric sequence**, where each term is derived from the previous term by multiplying it by \(\frac{2}{3}\).
The explicit formula for a geometric sequence can be written as:
\[
g(n) = g(0) \cdot r^n
\]
where \( r \) is the common ratio and \( g(0) \) is the initial term.
For our case:
- \( g(0) = 900 \)
- \( r = \frac{2}{3} \)
Therefore, the explicit formula becomes:
\[
g(n) = 900 \left( \frac{2}{3} \right)^n
\]
In conclusion, the answers are:
- The sequence is (Choice B) **Geometric sequence**.
- The explicit formula is:
\[
g(n) = 900 \left( \frac{2}{3} \right)^n
\]
Initially, the amount of contaminants is \( g(0) = 900 \) grams. Each hour, one-third of the contaminants is filtered out. This implies that two-thirds of the contaminants remain each hour.
Mathematically, we can express the amount of contaminants remaining at the beginning of the \( n^\text{th} \) hour as follows:
1. After the first hour, the amount remaining is:
\[
g(1) = \frac{2}{3} \cdot g(0) = \frac{2}{3} \cdot 900 = 600 \text{ grams}
\]
2. After the second hour:
\[
g(2) = \frac{2}{3} \cdot g(1) = \frac{2}{3} \cdot 600 = 400 \text{ grams}
\]
3. After the third hour:
\[
g(3) = \frac{2}{3} \cdot g(2) = \frac{2}{3} \cdot 400 \approx 266.67 \text{ grams}
\]
From this, we see that each term is produced by multiplying the previous term by a constant factor, \(\frac{2}{3}\). Thus, this sequence is a **geometric sequence**, where each term is derived from the previous term by multiplying it by \(\frac{2}{3}\).
The explicit formula for a geometric sequence can be written as:
\[
g(n) = g(0) \cdot r^n
\]
where \( r \) is the common ratio and \( g(0) \) is the initial term.
For our case:
- \( g(0) = 900 \)
- \( r = \frac{2}{3} \)
Therefore, the explicit formula becomes:
\[
g(n) = 900 \left( \frac{2}{3} \right)^n
\]
In conclusion, the answers are:
- The sequence is (Choice B) **Geometric sequence**.
- The explicit formula is:
\[
g(n) = 900 \left( \frac{2}{3} \right)^n
\]