Problem 20 - Contemporary Precalculus

Question

Show that the sequence is arithmetic and find its common difference. $$\left\\{-3-\frac{n}{2}\right\\}$$

Step-by-Step Solution

Answer

If so, what is the common difference? Answer: Yes, the sequence is arithmetic, and the common difference is 1/2.

Step 1: Find the n-th term and (n+1)-th term

The given sequence can be expressed as $$a_n = -3 - \frac{n}{2}$$ To find the (n+1)-th term, replace n with (n+1) in the expression for the n-th term: $$a_{n+1} = -3 - \frac{n+1}{2}$$

Step 2: Calculate the difference between the consecutive terms

Subtract the n-th term (\(a_n\)) from the (n+1)-th term (\(a_{n+1}\)): $$a_{n+1} - a_n = \left(-3 - \frac{n+1}{2}\right) - \left(-3 - \frac{n}{2}\right)$$

Step 3: Simplify the expression

Expanding and simplifying the expression obtained in Step 2: $$a_{n+1} - a_n = -3 - \frac{n+1}{2} + 3 + \frac{n}{2}$$ $$a_{n+1} - a_n = \frac{n}{2} - \frac{n}{2} + \frac{1}{2}$$ $$a_{n+1} - a_n = \frac{1}{2}$$

Step 4: Identify the common difference

Since \(a_{n+1} - a_n = \frac{1}{2}\) for all n, the difference between consecutive terms is constant. Therefore, the given sequence is arithmetic, and its common difference is: $$d = \frac{1}{2}$$

Key Concepts

Common Difference

When it comes to understanding arithmetic sequences, the concept of the 'common difference' is fundamental.

Imagine you are climbing stairs, and each step you take is the same height as the previous one. The 'common difference' in an arithmetic sequence functions in the same way; it's the distance, or difference, between any two consecutive terms. Mathematically, this common difference (\(d\)) is found by subtracting one term from the next term in the sequence.For instance, looking at our exercise's sequence \(a_n = -3 - \frac{n}{2}\), if we calculate the difference between two consecutive terms \(a_{n+1}\) and \(a_n\), we find a consistent difference of \(\frac{1}{2}\). This consistency confirms that the sequence is indeed arithmetic and \(\frac{1}{2}\) is our staircase's step height, metaphorically speaking. This constant step or common difference makes it easier to anticipate subsequent terms in the sequence simply by adding \(\frac{1}{2}\) to any given term.

n-th Term

The 'n-th term' is essentially the formula that represents the general term in an arithmetic sequence. It’s like a recipe that tells you how to make any term in the sequence, assuming you know which term number, or position, you’re interested in. The n-th term is expressed as:\[ a_n = a_1 + (n-1)d \]where \(a_1\) is the first term of the sequence, \(d\) is the common difference, and \(n\) denotes the term number. This formula is powerful because, with it, you can jump directly to any term in the sequence without having to list all the terms before it.In our exercise, you can see this in action: the n-th term is given by \(a_n = -3 - \frac{n}{2}\). By knowing any term number, you can determine the actual value of the term. For example, to find the fifth term, you simply plug \(n = 5\) into the formula, which would yield \(a_5 = -3 - \frac{5}{2}\). By using the n-th term formula, we bypass the need to write out the initial four terms, saving time and effort.

Consecutive Terms

To get more insight into arithmetic sequences, we need to examine 'consecutive terms.' These are terms that follow one after another, without any gaps between them, much like adjacent beads strung on a necklace.

In our sequence, \(a_n = -3 - \frac{n}{2}\) and \(a_{n+1} = -3 - \frac{n+1}{2}\) are consecutive terms. The importance of consecution lies in the pattern it creates; in an arithmetic sequence, every pair of consecutive terms has the same difference, which we call the common difference (\(d\)).

By calculating the difference between \(a_{n+1}\) and \(a_n\), we verify the sequence's arithmetic nature. The process we used in the exercise connects directly to how arithmetic sequences behave: regardless of the values of \(n\), the difference between any two consecutive terms is always \(\frac{1}{2}\), underscoring the sequence's predictability and uniformity, which is vital for various mathematical and real-world applications.