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An arithmetic sequence, also known as an arithmetic progression, is a series of numbers in which the difference between consecutive terms is constant. This constant difference is referred to as the 'common difference' and is denoted by the symbol 'd'.

For instance, in the sequence 2, 5, 8, 11, 14, ..., each term after the first is formed by adding the common difference of 3 to the preceding term. This property of adding a fixed number to go from one term to the next is the defining characteristic of an arithmetic sequence.

Understanding arithmetic sequences is fundamental because they're commonly found in various real-world scenarios, such as calculating simple interest over time or determining the total number of seats in an auditorium with incremental row lengths.

A sequence in mathematics is an ordered list of numbers that often follows a specific pattern or rule. The arithmetic sequence is one type of sequence. On the other hand, when we talk about series, we refer to the sum of the elements of a sequence.

For illustration, a sequence could be something like 3, 6, 9, 12, ..., while the series would be the sum of these numbers: 3 + 6 + 9 + 12 + ... . The series can be finite or infinite, depending on whether the sequence has a limited number of terms or goes on indefinitely.

Series have important applications in various fields, including finance for calculating annuities and in physics for analysing waveforms.

The 'nth term calculation' is a crucial concept when dealing with arithmetic sequences. It allows you to find any term in the sequence without having to list out all the preceding terms. The formula to find the nth term of an arithmetic sequence is given by: \[ a_n = a_1 + (n - 1)d \]Here, \(a_n\) represents the nth term you're looking for, \(a_1\) is the first term of the sequence, 'd' is the common difference, and 'n' is the term's position in the sequence.

This formula is instrumental in solving problems where you need to find the value of a term that's far along in the sequence, which would be impractical to do by counting one by one from the beginning. Using the nth term formula provided in our original problem, we swiftly calculated the fifteenth term, demonstrating the formula's efficiency and importance in handling arithmetic progressions.