- Identify the first term of the sequence as \(a_1 = -7.5\).
- Calculate the common difference as \(d = -9 - (-7.5) = -1.5\).
- Use the general formula for an arithmetic sequence \(a_n = a_1 + (n-1)d\) to get \(a_n = -7.5 + (n - 1)(-1.5)\).
- Conclude that the correct explicit formula is \(\boxed{a_n = -7.5 + (-1.5)(n - 1)}\).
Explanation
- Analyzing the Arithmetic Sequence
To find the explicit formula for the arithmetic sequence \(-7.5, -9, -10.5, -12, \ldots\), we first identify the components of an arithmetic sequence. An arithmetic sequence follows the general formula \(a_n = a_1 + (n-1)d\), where \(a_1\) is the first term and \(d\) is the common difference between consecutive terms. In our given sequence, the first term is \(a_1 = -7.5\).
- Calculating the Common Difference
Next, we calculate the common difference \(d\) by subtracting the first term from the second term. Given the terms \(a_1 = -7.5\) and \(a_2 = -9\), the common difference is:
\[d = a_2 - a_1 = -9 - (-7.5) = -9 + 7.5 = -1.5.\]
We can verify this with the next terms: \(a_3 - a_2 = -10.5 - (-9) = -1.5\), and \(a_4 - a_3 = -12 - (-10.5) = -1.5\). The constant common difference confirms that the sequence is indeed arithmetic with \(d = -1.5\).
- Determining the Explicit Formula
Now, we substitute the first term \(a_1 = -7.5\) and the common difference \(d = -1.5\) into the general formula \(a_n = a_1 + (n-1)d\):
\[a_n = -7.5 + (n - 1)(-1.5).\]
Rearranging this to match the structure of the options provided, we see:
\[a_n = -7.5 + (-1.5)(n - 1).\]
Comparing this to the given choices, it matches the first option.
Examples
Imagine a small business owner who monitors the depreciation of equipment. If a machine's value starts at \$7,500 and loses \$1,500 in value every year, its value over the years follows an arithmetic sequence. In our case, if the sequence represented annual value in thousands of dollars (scaled), the explicit formula would allow the owner to calculate the value of the machine in any given year \(n\) without having to list every single year's depreciation individually. This demonstrates how explicit formulas help in predicting future values or states in linear growth or decline processes.