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In arithmetic sequences, a recursion formula is used to describe the relationship between consecutive terms. Essentially, it provides a way to find any term in the sequence if the preceding term is known. For arithmetic sequences, the recursion formula can be expressed as:

• \( a_{n} = a_{n-1} + d \)

where \( a_{n} \) is the \( n \)-th term, \( a_{n-1} \) is the previous term, and \( d \) is the common difference.
This formula helps us understand how each term is constructed from the previous one, by simply adding the common difference to it. This makes it especially useful when you want to get the next few terms in a sequence without starting from the beginning each time. For example, if \( a_{1} = 5 \) and \( d = 6 \), then \( a_{2} = a_{1} + d = 5 + 6 = 11 \), confirming that the common difference is consistent throughout the sequence.

The common difference is a key characteristic of arithmetic sequences. It represents the constant amount by which consecutive terms increase or decrease. Identifying the common difference allows for the prediction of future terms in the sequence. To find the common difference, subtract any term from the subsequent term. For example, in the sequence described in the exercise, we have:

• \( a_2 - a_1 = 11 - 5 = 6 \)

This tells us that the common difference \( d \) is 6.
Understanding the common difference makes it much easier to forecast what the next numbers in the sequence will be without adding repeatedly. If you know this difference, you can jump to any term in the sequence through calculation rather than sequential construction.

The nth-term formula is a straightforward way to compute any term in an arithmetic sequence without having to list all the prior terms. It is configured to find any nth term directly through a formula based approach. The formula is:

• \( a_{n} = a_{1} + (n - 1) \cdot d \)

Where:

• \( a_{n} \) is the term you want to find.
• \( a_{1} \) is the first term of the sequence.
• \( n \) is the position of the term in the sequence.
• \( d \) is the common difference.

In the given exercise, to find \( a_{10} \), you apply the formula as such:
\( a_{10} = 5 + (10 - 1) \times 6 = 5 + 54 = 59 \).
This technique highlights the power of arithmetic progressions, allowing you to calculate terms independently and efficiently without prolonged sequential calculations.