- Identify the first term \(a_1 = \frac{3}{4}\).
- Calculate the common ratio by dividing the second term by the first term, \(r = \frac{1/4}{3/4} = \frac{1}{3}\).
- Use the general formula \(a_n = a_1 \cdot r^{(n-1)}\) to write the explicit formula.
- The final formula is \(\boxed{a_n = \frac{3}{4} \cdot \left( \frac{1}{3} \right)^{(n-1)}}\).
Explanation
- Analyzing the Given Sequence
The problem provides a sequence \(\frac{3}{4}, \frac{1}{4}, \frac{1}{12}, \frac{1}{36}, \ldots\) and states it is a geometric sequence. A geometric sequence takes the form \(a_n = a_1 \cdot r^{(n-1)}\), where \(a_1\) is the first term and \(r\) is the common ratio. Our task is to identify \(a_1\) and \(r\) to write the explicit formula for the sequence.
- Determining the First Term and Common Ratio
The first term of the sequence is explicitly given as \(a_1 = \frac{3}{4}\). The common ratio \(r\) can be determined by taking the ratio of any two consecutive terms. Using the first and second terms:
\[r = \frac{a_2}{a_1} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{4} \cdot \frac{4}{3} = \frac{1}{3}.\]
To verify this, we can check the third term \(a_3\):
\[a_3 = a_2 \cdot r = \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12}.\]
This matches the third term given in the sequence, confirming that \(r = \frac{1}{3}\).
- Assembling the Explicit Formula
Now that we have the first term \(a_1 = \frac{3}{4}\) and the common ratio \(r = \frac{1}{3}\), we can substitute these values into the general formula \(a_n = a_1 \cdot r^{(n-1)}\). The explicit formula for the sequence is:
\[a_n = \frac{3}{4} \cdot \left( \frac{1}{3} \right)^{(n-1)}.\]
This formula allows us to find any term in the sequence by inputting its position \(n\).
Examples
In real life, geometric sequences are often used to model scenarios where growth or decay occurs at a constant percentage rate. For example, if you have a piece of radioactive material that loses a constant fraction of its mass every day, or if you are modeling the amount of light passing through layers of tinted glass, you are using a geometric sequence. This formula helps predict the amount remaining after any number of steps \(n\), allowing for efficient planning and calculation in scientific or financial contexts.