- A geometric sequence is defined by a common ratio \(r\), such that each term \(a_n = a_{n-1} \cdot r\).
- The problem states each term is the previous term divided by \(2\), which is equivalent to multiplying by \(\frac{1}{2}\).
- Therefore, the common ratio is \(r = \frac{1}{2}\).
- The correct value is \(\boxed{\frac{1}{2}}\).
Explanation
- Understanding Geometric Sequences
The problem describes a sequence starting with the term \(a_1 = 12\). A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio, denoted by \(r\). The relationship between consecutive terms is given by \(a_n = a_{n-1} \cdot r\).
- Identifying the Common Ratio
We are given that each term in the sequence is the previous term divided by 2. This means that \(a_n = \frac{a_{n-1}}{2}\). We can rewrite this division as multiplication: \(a_n = a_{n-1} \cdot \frac{1}{2}\). Comparing this expression to the standard definition of a geometric sequence, \(a_n = a_{n-1} \cdot r\), we can identify the common ratio \(r\) as \(\frac{1}{2}\).
- Conclusion and Final Answer
The first term is \(a_1 = 12\). The second term is \(a_2 = 12 \cdot \frac{1}{2} = 6\). The third term is \(a_3 = 6 \cdot \frac{1}{2} = 3\), and so on. Since the rule is to multiply by \(\frac{1}{2}\) to get the next term, the common ratio is \(\frac{1}{2}\). Among the given options \(\frac{1}{2}\), \(2\), and \(12\), the correct value is \(\frac{1}{2}\).
Examples
In a real-life scenario, such as radioactive decay, a substance might lose half of its remaining mass in every time interval. If you start with 12 grams of a radioactive isotope that has a half-life of one hour, after one hour you will have 6 grams left, then 3 grams, and so on. The amount of substance remaining follows a geometric sequence where the common ratio is \(\frac{1}{2}\), allowing scientists to predict how long it will take for the substance to become safe.