Problem 68 - Precalculus with Limits

Question

In Exercises 67-70, write the first five terms of the sequence defined recursively. Use the pattern to write the \(n\)th term of the sequence as a function of \( n \). (Assume that \( n \) begins with 1.) \( a_1 = 25, a_{k + 1} = a_k - 5 \)

Step-by-Step Solution

Answer

The first five terms of the sequence are 25, 20, 15, 10, and 5. The \(n\)th term of the sequence as a function of \( n \) is \( a_n = 30 - 5n \).

Step 1: Compute the first five terms of the sequence

We start with the base of the sequence, \( a_1 = 25 \). Then, using the defined recursive formula, we calculate the following four terms. By plugging \( k = 1 \) into the formula, it gives \( a_2 = a_1 - 5 = 20 \). Likewise, we can continue to find more terms: \( a_3 = a_2 - 5 = 15 \), \( a_4 = a_3 - 5 = 10 \), \( a_5 = a_4 - 5 = 5 \). Therefore, the first five terms of the sequence are 25, 20, 15, 10, and 5.

Step 2: Derive the \(n\)th term of the sequence

In recursive formula, we are subtracting 5 from the previous term. So, we can observe that \( a_n = a_1 - 5(n - 1) \). The reason \( n-1 \) is used is because we start subtracting from the second term, not the first. Substituting \( a_1 = 25 \) into the formula we obtain \( a_n = 25 - 5(n - 1) = 30 - 5n \).

Key Concepts

Arithmetic Sequences

Arithmetic sequences are a type of number sequence where the difference between consecutive terms remains constant. This difference is known as the 'common difference.' In an arithmetic sequence, each term is derived by adding this common difference to the previous term. This pattern can be easily recognized by a steady increase or decrease in the sequence's numbers.

In the original exercise, we encountered an arithmetic sequence where the first term is 25, and each subsequent term decreases by 5. This makes the common difference -5. This specific type of sequence ensures that the first five terms are calculated by continually subtracting 5 from the previous term:

• The first term is given as 25.
• The second term becomes 25 - 5 = 20.
• Continuing this pattern, the other terms are 15, 10, and 5.

Understanding arithmetic sequences is crucial because they are foundational in mathematics, showing up in various mathematical contexts like calculus and series.

Sequence Formula

To express sequences mathematically, formulas are essential. They grant us the ability to determine any term in the sequence without listing all prior terms. In arithmetic sequences, there is a specific sequence formula often employed, which succinctly expresses terms as a function of their position, or 'n.'

For any arithmetic sequence, the formula to find the nth term is given by:
\[ a_n = a_1 + (n - 1)d \]
Where:

• \(a_n\) is the nth term we want to find.
• \(a_1\) is the first term of the sequence.
• \(d\) is the common difference.

Applying it to our exercise with \(a_1 = 25\) and \(d = -5\), we alter the formula:

• \(a_n = 25 - 5(n - 1)\)
• Simplifying further, \(a_n = 30 - 5n\)

This finding confirms that the sequence does decrease by 5 at each step efficiently, and gives us insight on how jumps or skips in the sequence impact the final outcome.

Recursive Formula

Recursive formulas are different from explicit formulas as they define each term using the previous term(s), establishing a formula as a step-by-step process. In sequences, using a recursive formula can sometimes provide an easier manner to generate terms, especially when patterns are present.

In the given exercise, the recursive formula is:

• \(a_1 = 25\)
• \(a_{k+1} = a_k - 5\)

This tells us that to find any subsequent term, you subtract 5 from the previous term. Recursive formulas don't typically give a direct calculation for any specific term, unlike the sequence formula, but they excel in clarifying the relationship from one term to the next.
When approaching recursive sequences:

• Start with the known initial term(s) like \(a_1 = 25\).
• Use the recursive relation to find further terms.

As a powerful tool in mathematics, recursive formulas elegantly describe processes over iterations, natural in computational contexts.