Problem 68 - Precalculus Mathematics for Calculus

Question

The arithmetic mean (or average) of two numbers \(a\) and \(b\) is $$ m=\frac{a+b}{2} $$ Note that \(m\) is the same distance from \(a\) as from \(b,\) so \(a, m, b\) is an arithmetic sequence. In general, if \(m_{1}, m_{2}, \ldots, m_{k}\) are equally spaced between \(a\) and \(b\) so that $$ a, m_{1}, m_{2}, \dots, m_{k}, b $$ is an arithmetic sequence, then \(m_{1}, m_{2}, \ldots, m_{k}\) are called \(k\) arithmetic means between \(a\) and \(b\). (a) Insert two arithmetic means between 10 and 18 . (b) Insert three arithmetic means between 10 and 18 . (c) Suppose a doctor needs to increase a patient's dosage of a certain medicine from 100 mg to 300 mg per day in five equal steps. How many arithmetic means must be inserted between 100 and 300 to give the progression of daily doses, and what are these means?

Step-by-Step Solution

Answer

(a) Two means: \(\frac{38}{3}, \frac{46}{3}\) (b) Three means: 12, 14, 16 (c) Means: 140 mg, 180 mg, 220 mg, 260 mg.

Step 1: Understand the Arithmetic Sequence

An arithmetic sequence is a sequence of numbers in which the difference between consecutive terms is constant. Let's denote the first term by \(a\), the last term by \(b\), and the number of arithmetic means by \(k\). The common difference \(d\) of the sequence is given by \(d = \frac{b-a}{k+1}\).

Step 2: Solve Part (a) – Insert Two Arithmetic Means

We need to insert two arithmetic means between 10 and 18, meaning \(k = 2\). Calculate the common difference \(d\) with \(d = \frac{18 - 10}{2 + 1} = \frac{8}{3} = \frac{8}{3}\). Now, find the means: \(m_1 = 10 + \frac{8}{3} = \frac{38}{3}\), and \(m_2 = 10 + 2 \times \frac{8}{3} = \frac{46}{3}\). The sequence is \(10, \frac{38}{3}, \frac{46}{3}, 18\).

Step 3: Solve Part (b) – Insert Three Arithmetic Means

Inserting three arithmetic means between 10 and 18 means \(k = 3\). Calculate the common difference \(d\) using \(d = \frac{18 - 10}{3 + 1} = 2\). The arithmetic means are: \(m_1 = 12\), \(m_2 = 14\), and \(m_3 = 16\). The sequence is \(10, 12, 14, 16, 18\).

Step 4: Solve Part (c) – Increase Dosage from 100 mg to 300 mg

The range is from 100 mg to 300 mg in five equal steps. So we need \(5 - 1 = 4\) steps between doses. Calculate the common difference \(d\): \(d = \frac{300 - 100}{5} = 40\). The means are: 140 mg, 180 mg, 220 mg, and 260 mg. The complete sequence is 100 mg, 140 mg, 180 mg, 220 mg, 260 mg, 300 mg.

Key Concepts

Understanding Arithmetic Mean

The arithmetic mean, often called the average, is a fundamental concept in mathematics used to find the central value of a set of numbers. To compute the arithmetic mean of two numbers, simply add the numbers together and divide by two. The formula is \( m = \frac{a+b}{2} \). This mean represents a value that is equidistant from either number, forming a connection in arithmetic sequences. This concept is useful not only as a measure of central tendency but also in sequence analysis, where inserting arithmetic means can form new sequences. For example, inserting numbers equally spaced between given values.

The Importance of Common Difference

In an arithmetic sequence, the common difference \(d\) is what makes the sequence unique. It represents the regular interval between consecutive terms in the sequence. Calculating \(d\) requires subtracting the previous term from the current term or using the initial and final terms and the number of intervals. The formula \(d = \frac{b-a}{k+1}\) helps in determining this difference when inserting additional terms, called arithmetic means, between two numbers. Understanding and applying the common difference is crucial in effectively managing number sequences and solving sequence-related problems.

The Sequence Progression and its Properties

Sequence progression in arithmetic is characterized by having equal differences between consecutive terms. Progressions, such as those used in coursework or real-world applications like medicine dosages, require knowledge of arithmetic sequences. The main property is the fixed increment \(d\). This can create predictable patterns, which are beneficial in various mathematical applications. Arithmetic progression can be smoothly expanded to include any number of intermediate terms, allowing one to analyze or predict outcomes based on existing data.

Mathematics Problem-Solving Techniques

Solving problems involving arithmetic sequences necessitates logical steps and clear methods. Identifying all given values, such as initial and final terms and the number of desired means, is the first step. Applying known formulas for arithmetic progression aids in finding the common difference and subsequently the intermediary terms. Problem-solving becomes intuitive when following systematic approaches:

• Identify given values and requirements.
• Calculate common differences using the correct formula.
• Apply consistent sequences to fill in missing terms.

Mastery of these techniques supports not only academic tasks but also real-life applications, making them a vital part of any math curriculum.