Problem 41 - Algebra and Trigonometry Real Mathematics, Real People

Question

Write the first five terms of the arithmetic sequence. Find the common difference and write the \(n\) th term of the sequence as a function of \(n .\) $$a_{1}=\frac{3}{5}, a_{k+1}=-\frac{1}{10}+a_{k}$$

Step-by-Step Solution

Answer

The first five terms of the arithmetic sequence are: \(\frac{3}{5}, \frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \frac{1}{5}\). The common difference, d, is \(-\frac{1}{10}\). Lastly, the nth term of the sequence as a function of n based on the solution steps is: \(a_n = \frac{3}{5} + (n-1) * (-\frac{1}{10})\).

Step 1: Calculate the first five terms

Calculate the first five terms of the sequence using the formula \(a_{k+1} = -\frac{1}{10} + a_k\). Begin with the first term provided ( \(a_1 = \frac{3}{5}\)). Then, use the formula recursively to calculate terms two to five: For \(a_2\), use \(a_1\) in the formula, obtaining \(a_2 = -\frac{1}{10} + a_1\), and repeat this process for \(a_3, a_4, a_5\).

Step 2: Find the common difference

In an arithmetic progression, the common difference is the difference between two successive terms. This is what you get when you subtract \(a_k\) from \(a_{k+1}\), any terms in the sequence. In the given sequence, take \(a_2\) and subtract \(a_1\). The result is the common difference. Repeat this with \(a_3\) and \(a_2\), \(a_4\) and \(a_3\), and so on to verify that the difference remains consistent.

Step 3: Write the nth term as a function of n

The general formula for the nth term of an arithmetic sequence is \(a_n = a_1 + (n-1) * d\), where \(d\) is the common difference. Now, replace \(a_1\) with the given initial value and \(d\) with the common difference obtained in Step 2. You will then have the nth term of the sequence as a function of n.

Key Concepts

Common Difference

An arithmetic sequence is a sequence of numbers with a constant difference between consecutive terms. This constant is known as the "common difference." It is denoted by the letter \(d\). To find the common difference, you subtract any term from the next term in the sequence.
For instance, in our exercise, the common difference is calculated by taking the second term \(a_2\) and subtracting the first term \(a_1\). This procedure ensures that the common difference is the same throughout the sequence.

• If \(a_2\) is \(-\frac{1}{10} + a_1\), then you can subtract \(a_1\) from \(a_2\) to find \(d\).
• This difference \(d\) is crucial as it defines how the sequence progresses.

Consistency of this difference confirms the sequence is arithmetic, which simplifies predicting future terms.

Recursive Formula

The recursive formula in an arithmetic sequence allows you to generate terms of the sequence from previous terms. In mathematical terms, a recursive formula is an equation that relates each term of a sequence to some of its predecessors.
In the provided exercise, the recursive formula is \(a_{k+1} = -\frac{1}{10} + a_k\). This means each term is obtained by adding \(-\frac{1}{10}\) to the preceding term.
Important aspects of recursive formulas include:

• They often require you to know one or more initial terms. In our exercise, you begin with \(a_1 = \frac{3}{5}\).
• You progressively use the formula to calculate subsequent terms. For example, given \(a_1\), you can find \(a_2\), then \(a_3\), and so forth.
• This approach highlights the relationship between terms, emphasizing the sequence's structure.

Recursive formulas provide a simple method to understand sequences one step at a time, especially useful when manually finding the first few terms.

Nth Term Formula

The \(n\)th term formula allows you to calculate any term in the sequence directly without having to go through all the previous terms. It is a very useful formula when you need to find an arbitrarily distant term in the sequence.
The general form for the \(n\)th term of an arithmetic sequence is \(a_n = a_1 + (n-1) \cdot d\). In this formula:

• \(a_1\) represents the first term of the sequence.
• \(n\) stands for the term number you wish to find.
• \(d\) is the common difference, found in our first step.

In the context of our exercise, after finding \(d\), you replace \(a_1\) with \(\frac{3}{5}\) and use the value of \(d\) to write \(a_n\) as an explicit function of \(n\).
This formula is particularly advantageous as it allows you to quickly calculate not just the first few terms, but any term in the sequence, offering convenience and great insight into the sequence's behavior.