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Understanding the common difference is crucial for working with arithmetic sequences. It is the difference between any two consecutive terms in the sequence. To find it, simply subtract one term from the next. For example, in a sequence where the first term is 5 and the fourth term is 15, you can calculate the common difference by subtracting the first term from the fourth term, and then dividing by the number of steps between them, which is 3 in this case.

This calculation results in the common difference, denoted as 'd', which is \( \frac{15 - 5}{4 - 1} = \frac{10}{3} \). Once identified, this constant difference is used to find subsequent terms and can be seen as the heartbeat of the sequence, pulsing at a regular interval. In our case, every term of the sequence grows by \( \frac{10}{3} \) from the previous term.

The nth term of an arithmetic sequence is determined using the formula \( a_n = a_1 + (n - 1) \cdot d \). Here, \( a_1 \) is the first term, \((n - 1)\) signifies the number of terms away from the first term that we are looking to find, and 'd' represents the common difference.

For the given sequence, we plug the values into this formula: the first term \( a_1 = 5 \) and the common difference \( d = \frac{10}{3} \). By substituting these values, we develop a specific formula for our particular sequence, which enables us to compute any term within the sequence effortlessly.

Thus, for any integer 'n', the nth term can be computed by \( a_n = 5 + (n - 1) \cdot \frac{10}{3} \), which is an essential concept for evaluating terms at any position in our arithmetic sequence.

The process of sequence simplification involves reformatting the nth term formula to make it cleaner and often easier to work with. After applying the general nth term formula to our particular sequence, our objective is to simplify the expression so that it can be more practically used and understood.

By expanding and combining like terms, we arrive at a simpler expression. In our sequence's formula, we distribute the common difference across the \((n - 1)\) term, which initially gives us \( 5 + \frac{10n - 10}{3} \). Simplifying further, we combine the constant terms and adjust the numerator to obtain the simplified nth term formula \( \frac{10n + 5}{3} \).

This final simplified form for the nth term, \( a_n = \frac{10n + 5}{3} \), takes away the distraction of dealing with multiple steps and makes it more straightforward to compute any term in the sequence, facilitating an increased focus on the arithmetic progression itself.