Arithmetic Sequences And Means Practice Exercise 3.1 Solutions

This article provides a comprehensive walkthrough of Practice Exercise 3.1, focusing on arithmetic sequences and arithmetic means. We will dissect each problem step-by-step, ensuring a clear understanding of the underlying concepts. Whether you're a student tackling homework or simply looking to brush up on your math skills, this guide offers detailed explanations and solutions.

1. Understanding Arithmetic Sequences

The core concept here is the arithmetic sequence, a sequence of numbers where the difference between consecutive terms remains constant. This constant difference is known as the common difference. Identifying this common difference is crucial for solving problems related to arithmetic sequences.

1. a. Writing a Formula for the nth Term

Arithmetic sequences are fundamental in mathematics, and understanding how to derive a formula for the nth term is crucial for solving various problems. In this specific sequence, 9, 13, 17, we observe a consistent pattern. To begin, we need to identify the common difference, which is the constant value added to each term to get the next term in the sequence. By subtracting the first term from the second term (13 - 9), we find that the common difference, often denoted as d, is 4. Similarly, subtracting the second term from the third term (17 - 13) also yields 4, confirming that this is indeed an arithmetic sequence. The first term of the sequence, denoted as a₁, is 9.

The general formula for the nth term (aₙ) of an arithmetic sequence is given by: aₙ = a₁ + (n - 1)d. This formula allows us to calculate any term in the sequence without having to list out all the preceding terms. It's a powerful tool for solving problems and understanding the behavior of arithmetic sequences. Now, we can substitute the values we've identified into this formula. Here, a₁ is 9 and d is 4. Plugging these values into the general formula, we get: aₙ = 9 + (n - 1)4. This equation represents the nth term of the given arithmetic sequence. To simplify this formula, we can distribute the 4 across the terms inside the parentheses: aₙ = 9 + 4n - 4. Then, combining the constant terms (9 and -4), we arrive at the simplified formula: aₙ = 4n + 5. This simplified formula provides a direct way to calculate any term in the sequence by simply substituting the term number (n) into the equation. For instance, to find the 10th term, we would substitute n = 10 into the formula. This formula is not only useful for finding specific terms but also for understanding the sequence's overall structure and how it progresses.

1. b. Finding the 12th Term

Using the formula derived in the previous step, finding the 12th term becomes straightforward. The 12th term of the sequence is found by substituting n = 12 into the formula aₙ = 4n + 5. This involves replacing the variable n with the number 12, which represents the term number we are interested in. So, we have: a₁₂ = 4(12) + 5. Now, we perform the arithmetic operations. First, multiply 4 by 12, which gives us 48. Then, add 5 to the result: a₁₂ = 48 + 5. This simple addition leads us to the final answer: a₁₂ = 53. Therefore, the 12th term of the arithmetic sequence is 53. This means that if we were to continue the sequence 9, 13, 17, ..., the 12th number in the sequence would be 53. This calculation demonstrates the power and efficiency of the formula aₙ = 4n + 5. It allows us to quickly determine any term in the sequence without having to manually list out all the preceding terms. This is particularly useful for finding terms that are far out in the sequence, where manual calculation would be time-consuming and prone to error. Furthermore, understanding how to use this formula is a fundamental skill in dealing with arithmetic sequences and is applicable in various mathematical contexts and real-world scenarios involving patterns and progressions.

1. c. Determining Which Term is 79

To determine which term in the sequence is equal to 79, we again utilize the formula aₙ = 4n + 5. However, in this case, we are given the value of aₙ (which is 79) and need to find the value of n, which represents the term number. This is a reverse application of the formula compared to the previous step, but it's equally important for understanding arithmetic sequences. To find the term number, we set aₙ equal to 79 and solve for n. The equation becomes: 79 = 4n + 5. Now, we need to isolate n on one side of the equation. First, we subtract 5 from both sides of the equation to eliminate the constant term on the right side: 79 - 5 = 4n + 5 - 5. This simplifies to 74 = 4n. Next, to solve for n, we divide both sides of the equation by 4: 74 / 4 = 4n / 4. This gives us n = 18.5. However, since n represents the term number in a sequence, it must be a whole number. The fact that we obtained a non-integer value indicates that 79 is not a term in this arithmetic sequence. This is an important observation and highlights the nature of arithmetic sequences: not every number will be a term in a given sequence. The terms are determined by the common difference and the first term, and only numbers that fit the pattern will be included. If the result had been a whole number, say 19, then 79 would have been the 19th term in the sequence. This process of solving for n demonstrates the versatility of the nth term formula and its ability to address different types of questions about arithmetic sequences.

2. Inserting Arithmetic Means

Arithmetic means are the terms inserted between two given terms in an arithmetic sequence. The goal is to create a continuous arithmetic progression. This process involves finding the common difference that fits the given terms and the number of means to be inserted. Inserting arithmetic means is a fundamental concept in understanding and manipulating arithmetic sequences, as it allows us to create a smooth transition between two numbers while maintaining the constant difference characteristic of these sequences.

To insert three arithmetic means between 11 and 39, we need to create an arithmetic sequence where 11 is the first term (a₁) and 39 is the fifth term (a₅), as inserting three terms between two numbers results in a total of five terms. The problem essentially asks us to find three numbers that, when placed between 11 and 39, form an arithmetic sequence. This means that the difference between consecutive terms must be constant throughout the sequence. To solve this, we first need to determine the common difference (d) of the sequence. We know that a₁ = 11 and a₅ = 39. Using the formula for the nth term of an arithmetic sequence, aₙ = a₁ + (n - 1)d, we can substitute the known values to solve for d. In this case, we substitute n = 5, a₅ = 39, and a₁ = 11 into the formula: 39 = 11 + (5 - 1)d. Simplifying this equation, we get 39 = 11 + 4d. Now, we need to isolate d to find its value. Subtract 11 from both sides of the equation: 39 - 11 = 4d, which simplifies to 28 = 4d. Then, divide both sides by 4: 28 / 4 = d, which gives us d = 7. Now that we have the common difference, we can find the three arithmetic means by successively adding d to the previous term, starting with the first term, 11. The first arithmetic mean (a₂) is 11 + 7 = 18. The second arithmetic mean (a₃) is 18 + 7 = 25. And the third arithmetic mean (a₄) is 25 + 7 = 32. Thus, the three arithmetic means between 11 and 39 are 18, 25, and 32. This means the complete arithmetic sequence is 11, 18, 25, 32, 39. This exercise demonstrates the application of the nth term formula in a slightly different context, highlighting its versatility in solving problems related to arithmetic sequences.

3. Inserting Arithmetic Means with Algebraic Expressions

This problem extends the concept of inserting arithmetic means to include algebraic expressions. The approach remains the same, but now we are dealing with variables, which adds a layer of complexity. The key is to apply the same principles of arithmetic sequences while carefully handling the algebraic manipulations. This type of problem not only tests the understanding of arithmetic sequences but also reinforces algebraic skills.

To insert two arithmetic means between x - 4 and 4x + 5, we follow a similar approach as in the previous problem, but this time, we're dealing with algebraic expressions instead of numerical values. This means the arithmetic means will also be expressed in terms of x. The first term, a₁, is given as x - 4, and the fourth term, a₄, is 4x + 5, since inserting two means creates a sequence with four terms in total. We need to find two expressions that fit between these terms to form an arithmetic sequence. This involves finding the common difference (d) in terms of x. We use the same formula for the nth term of an arithmetic sequence: aₙ = a₁ + (n - 1)d. Substituting the known values, where n = 4, a₁ = x - 4, and a₄ = 4x + 5, we get: 4x + 5 = (x - 4) + (4 - 1)d. This simplifies to 4x + 5 = x - 4 + 3d. Now, we solve for d. First, we combine like terms and isolate the term with d. Subtract x from both sides: 4x + 5 - x = x - 4 + 3d - x, which simplifies to 3x + 5 = -4 + 3d. Then, add 4 to both sides: 3x + 5 + 4 = -4 + 3d + 4, which simplifies to 3x + 9 = 3d. Finally, divide both sides by 3 to solve for d: (3x + 9) / 3 = 3d / 3, which gives us d = x + 3. Now that we have the common difference, we can find the two arithmetic means. The second term, a₂, is found by adding d to a₁: a₂ = (x - 4) + (x + 3) = 2x - 1. The third term, a₃, is found by adding d to a₂: a₃ = (2x - 1) + (x + 3) = 3x + 2. Therefore, the two arithmetic means between x - 4 and 4x + 5 are 2x - 1 and 3x + 2. This completes the arithmetic sequence: x - 4, 2x - 1, 3x + 2, 4x + 5. This problem effectively demonstrates how the principles of arithmetic sequences can be applied even when dealing with algebraic expressions, requiring careful algebraic manipulation and a solid understanding of the underlying concepts.

This comprehensive guide has walked you through Practice Exercise 3.1, covering key concepts related to arithmetic sequences and means. By understanding the formulas and techniques discussed, you'll be well-equipped to tackle similar problems in the future. Remember, practice is key to mastering these concepts, so keep working on various examples to solidify your understanding.