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In arithmetic sequences, finding the nth term is a critical skill in understanding how the sequence is structured. To achieve this, there's a handy formula you can use: \[a_n = a_1 + (n - 1) \times d\] Here, \(a_n\) is the nth term that you're interested in, \(a_1\) is the first term in the sequence, \(n\) is the position of the term you're looking for, and \(d\) is the common difference between each consecutive term. In our exercise, the first term \(a_1\) is given as 100, and the common difference \(d\) is -8. By substituting these values into the nth term formula, we can find any term in the series. Learning to apply this formula is fundamental for solving any arithmetic sequence problem.

Simplifying arithmetic sequences helps in verifying the correct formula and spotting any mistakes. After substituting the known values into the nth term formula, you often need to simplify the expression to make it more workable. For instance, starting with: \[a_n = 100 + (n - 1) \times (-8)\] The next step involves distributing the multiplication across the terms: 1. Multiply \( (n-1) \times -8\) to get \(-8n + 8\). 2. Replace the expression in the equation to get \(a_n = 100 - 8n + 8\). 3. Finally, combine like terms and simplify: \[a_n = 108 - 8n\] By simplifying, the nth term equation becomes easier to use and understand.

Arithmetic Progression (or sequence) is a sequence of numbers in which the difference between consecutive terms is constant. This difference is known as the common difference, \(d\). Understanding arithmetic progressions is key to mastering sequences and series in mathematics. In the given exercise, we deal with an arithmetic sequence where every term is derived by subtracting 8 from the previous one. This is signified by \(d = -8\). The pattern of the sequence can be broken down as: - Start with the first term: 100. - Subtract 8 to get the next term: 92. - Continue this process to observe the progression: - Next terms are 84, 76, 68, and so forth. Each of these steps highlights the repetitive and predictable nature of arithmetic progressions. Recognizing these patterns allows you to use them efficiently in calculations.