7- Sequences

# 7- Sequences ## Definition and types of sequences, Arithmetic progression: General term and the arithmetic mean ### Introduction Welcome to our lesson on sequences! In mathematics, a sequence is simply an ordered list of numbers. We encounter sequences in various real-life situations, from predicting population growth to calculating interest on savings. In this lesson, we'll start by understanding what sequences are and explore some basic types. Then, we will dive into a very important type of sequence called an Arithmetic Progression (AP), learn how to find any term in an AP, and understand the concept of the arithmetic mean. ### Key Concepts #### What is a Sequence? * A sequence is an ordered list of numbers or objects that follow a specific pattern or rule. * Each number in the sequence is called a **term**. * Terms are usually separated by commas. * **Example:** 2, 4, 6, 8, ... (This sequence of even numbers) #### Notation * The terms of a sequence are typically denoted using a subscript: \( a_1, a_2, a_3, ext{...}, a_n \). * \( a_1 \) represents the first term. * \( a_2 \) represents the second term. * \( a_n \) represents the \( n \)-th term, also known as the general term. #### Types of Sequences * **Finite Sequence:** A sequence that has a limited number of terms. It has a definite beginning and a definite end. * **Example:** 1, 3, 5, 7 (This sequence has 4 terms). * **Infinite Sequence:** A sequence that continues indefinitely without an end. It is often indicated by an ellipsis (...). * **Example:** 2, 4, 6, 8, ... (This sequence continues forever). * **Convergent Sequence:** A sequence whose terms get closer and closer to a specific finite value as \( n \) gets larger. * **Divergent Sequence:** A sequence whose terms do not approach a specific finite value (they might grow infinitely large, infinitely small, or oscillate). #### Arithmetic Progression (AP) ##### Definition of an Arithmetic Progression * An **Arithmetic Progression (AP)** is a sequence in which the difference between consecutive terms is constant. * This constant difference is called the **common difference**, denoted by \( d \). * To get any term from the preceding term, you simply add the common difference \( d \). * **Example:** 3, 7, 11, 15, ... * Here, \( 7-3 = 4 \), \( 11-7 = 4 \), \( 15-11 = 4 \). * The common difference \( d \) is 4. ##### Finding the Common Difference (d) * The common difference \( d \) can be found by subtracting any term from its succeeding term: \[ d = a_2 - a_1 = a_3 - a_2 = ext{...} = a_n - a_{n-1} \] ##### The General Term (nth Term) of an AP * If \( a \) (or \( a_1 \)) is the first term and \( d \) is the common difference, then the terms of an AP are: * \( a_1 = a \) * \( a_2 = a + d \) * \( a_3 = a + 2d \) * \( a_4 = a + 3d \) * And so on... * The general formula for the \( n \)-th term of an Arithmetic Progression is: \[ a_n = a + (n-1)d \] Where: * \( a_n \) is the \( n \)-th term * \( a \) (or \( a_1 \)) is the first term * \( n \) is the term number (position of the term in the sequence) * \( d \) is the common difference ##### The Arithmetic Mean * If three numbers \( x, y, z \) are in an Arithmetic Progression, then \( y \) is called the **arithmetic mean** of \( x \) and \( z \). * This means the common difference between \( x \) and \( y \) is the same as between \( y \) and \( z \): \[ y - x = z - y \] * Rearranging this equation to find \( y \): \[ 2y = x + z \] \[ y = rac{x+z}{2} \] * Therefore, the arithmetic mean of any two numbers \( A \) and \( B \) is simply their sum divided by two: \( rac{A+B}{2} \). ### Vocabulary * **Sequence:** An ordered list of numbers. * **Term:** Each number in a sequence. * **Finite Sequence:** A sequence with a limited number of terms. * **Infinite Sequence:** A sequence that continues indefinitely. * **Arithmetic Progression (AP):** A sequence where the difference between consecutive terms is constant. * **Common Difference (d):** The constant difference between consecutive terms in an AP. * **General Term (nth term):** A formula to find any term in a sequence, usually denoted by \( a_n \). * **Arithmetic Mean:** For two numbers, it is their sum divided by two. It is the middle term if three terms form an AP. ### Examples #### Example 1: Identifying Sequences and Types Identify if the following are sequences, and if they are, state their type (finite/infinite) and if they are an AP. If an AP, state the common difference. a) 1, 5, 9, 13, 17 b) 1, 2, 4, 8, ... c) 10, 7, 4, 1, ... **Solution:** a) Yes, this is a finite sequence. \( 5-1=4, 9-5=4, ext{...} \). It is an AP with \( d=4 \). b) Yes, this is an infinite sequence. \( 2/1=2, 4/2=2, ext{...} \). It is not an AP (it's a Geometric Progression). c) Yes, this is an infinite sequence. \( 7-10 = -3, 4-7 = -3, ext{...} \). It is an AP with \( d=-3 \). #### Example 2: Finding the nth Term Find the 10th term of the AP: 3, 7, 11, ... **Solution:** First term \( a = 3 \). Common difference \( d = 7 - 3 = 4 \). We want to find the 10th term, so \( n = 10 \). Using the formula \( a_n = a + (n-1)d \): \[ a_{10} = 3 + (10-1)4 \] \[ a_{10} = 3 + (9)4 \] \[ a_{10} = 3 + 36 \] \[ a_{10} = 39 \] The 10th term is 39. #### Example 3: Finding the Position of a Term Which term of the AP 5, 8, 11, ... is 32? **Solution:** First term \( a = 5 \). Common difference \( d = 8 - 5 = 3 \). We are given \( a_n = 32 \) and need to find \( n \). Using the formula \( a_n = a + (n-1)d \): \[ 32 = 5 + (n-1)3 \] Subtract 5 from both sides: \[ 32 - 5 = (n-1)3 \] \[ 27 = 3(n-1) \] Divide by 3: \[ rac{27}{3} = n-1 \] \[ 9 = n-1 \] Add 1 to both sides: \[ n = 9 + 1 \] \[ n = 10 \] So, 32 is the 10th term of the AP. #### Example 4: Arithmetic Mean Find the arithmetic mean of 12 and 20. **Solution:** Using the formula \( y = rac{x+z}{2} \): \[ y = rac{12+20}{2} \] \[ y = rac{32}{2} \] \[ y = 16 \] The arithmetic mean of 12 and 20 is 16. ### Exercises 1. For each sequence, determine if it is an Arithmetic Progression. If it is, find the common difference \( d \) and the next two terms. a) 10, 8, 6, 4, ... b) 1, 3, 9, 27, ... c) -5, 0, 5, 10, ... 2. Find the 15th term of the AP: 1, 6, 11, ... 3. The first term of an AP is 7 and the common difference is 3. Find the 8th term. 4. If the \( n \)-th term of an AP is given by \( a_n = 2n + 5 \), find the first three terms of the sequence. 5. Which term of the AP 2, 9, 16, ... is 72? 6. Insert three arithmetic means between 10 and 30. (This means forming an AP like: 10, AM1, AM2, AM3, 30). ### Assignment 1. The third term of an AP is 10 and the seventh term is 26. Find the first term \( a \) and the common difference \( d \) of the AP. 2. An AP has a first term of -5 and a common difference of 4. Is 50 a term in this sequence? Justify your answer. 3. If \( 2x, x+3, 3x-1 \) are consecutive terms of an AP, find the value of \( x \). 4. The 11th term of an AP is 39 and the 21st term is 79. Find the 1st term and the common difference. ### Summary * A **sequence** is an ordered list of numbers following a pattern, where each number is a **term**. * Sequences can be **finite** (limited terms) or **infinite** (continues indefinitely). * An **Arithmetic Progression (AP)** is a special type of sequence where the difference between consecutive terms is constant. This constant difference is called the **common difference (\( d \))**. * The formula for the **\( n \)-th term** of an AP is \( a_n = a + (n-1)d \), where \( a \) is the first term, \( n \) is the term number, and \( d \) is the common difference. * The **arithmetic mean** of two numbers \( A \) and \( B \) is \( rac{A+B}{2} \). If three terms are in AP, the middle term is the arithmetic mean of the other two. ## Sum of the first n terms of an AP, Gp,  General term and the geometric mean ### Introduction Welcome to this lesson on sequences, a fundamental topic in mathematics! Today, we will dive into two special types of sequences: Arithmetic Progressions (AP) and Geometric Progressions (GP). We'll learn how to find any term in these sequences, calculate the sum of their first few terms, and understand the concept of the geometric mean. These concepts are crucial for solving various real-world problems, from finance to physics. ### Key Concepts #### 1. Arithmetic Progression (AP) An Arithmetic Progression (AP) is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the **common difference**. ##### General Term (\(n^{th}\) term) of an AP If the first term of an AP is \(a\) and the common difference is \(d\), then the \(n^{th}\) term, denoted by \(T_n\) or \(a_n\), is given by the formula: \[ T_n = a + (n-1)d \] * \(T_n\) = the \(n^{th}\) term * \(a\) = the first term * \(n\) = the term number * \(d\) = the common difference ##### Sum of the first \(n\) terms of an AP The sum of the first \(n\) terms of an AP, denoted by \(S_n\), can be calculated using two main formulas: 1. If the first term (\(a\)) and the common difference (\(d\)) are known: \[ S_n = \frac{n}{2} [2a + (n-1)d] \] 2. If the first term (\(a\)) and the last term (\(l\) or \(T_n\)) are known: \[ S_n = \frac{n}{2} (a + l) \] #### 2. Geometric Progression (GP) A Geometric Progression (GP) is a sequence of non-zero numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the **common ratio**. ##### General Term (\(n^{th}\) term) of a GP If the first term of a GP is \(a\) and the common ratio is \(r\), then the \(n^{th}\) term, denoted by \(T_n\) or \(a_n\), is given by the formula: \[ T_n = ar^{n-1} \] * \(T_n\) = the \(n^{th}\) term * \(a\) = the first term * \(n\) = the term number * \(r\) = the common ratio ##### Sum of the first \(n\) terms of a GP The sum of the first \(n\) terms of a GP, denoted by \(S_n\), is given by: 1. If the common ratio \(r \neq 1\): \[ S_n = \frac{a(r^n - 1)}{r - 1} \quad \text{if } r > 1 \text{ or } r < -1 \] or \[ S_n = \frac{a(1 - r^n)}{1 - r} \quad \text{if } -1 < r < 1 \] 2. If the common ratio \(r = 1\): \[ S_n = na \] #### 3. Geometric Mean (GM) The Geometric Mean (GM) for two positive numbers \(a\) and \(b\) is the square root of their product. It is a type of average that is useful for sets of positive numbers that are interpreted as rates of change, growth, or percentages. ##### Formula for Geometric Mean If \(G\) is the geometric mean between two positive numbers \(a\) and \(b\), then: \[ G = \sqrt{ab} \] It also implies that \(a, G, b\) form a geometric progression, so \( \frac{G}{a} = \frac{b}{G} \implies G^2 = ab \). ### Vocabulary * **Sequence:** An ordered list of numbers. * **Series:** The sum of the terms of a sequence. * **Arithmetic Progression (AP):** A sequence where the difference between consecutive terms is constant. * **Common Difference (d):** The constant difference between consecutive terms in an AP. * **Geometric Progression (GP):** A sequence where the ratio between consecutive terms is constant. * **Common Ratio (r):** The constant ratio between consecutive terms in a GP. * **General Term (n-th term):** A formula that allows you to find any term in a sequence given its position \(n\). * **Sum of n terms:** The total obtained by adding the first \(n\) terms of a sequence. * **Geometric Mean (GM):** For two positive numbers, the square root of their product, forming a GP with them. ### Examples #### Example 1: Finding the General Term and Sum of an AP Consider the AP: 3, 7, 11, 15, ... a) Find the \(12^{th}\) term. b) Find the sum of the first 10 terms. **Solution:** Here, \(a = 3\) and \(d = 7 - 3 = 4\). a) For the \(12^{th}\) term (\(n=12\)): \[ T_{12} = a + (12-1)d = 3 + (11)(4) = 3 + 44 = 47 \] The \(12^{th}\) term is 47. b) For the sum of the first 10 terms (\(n=10\)): \[ S_{10} = \frac{10}{2} [2(3) + (10-1)4] = 5 [6 + (9)(4)] = 5 [6 + 36] = 5 [42] = 210 \] The sum of the first 10 terms is 210. #### Example 2: Finding the General Term and Sum of a GP Consider the GP: 2, 6, 18, 54, ... a) Find the \(6^{th}\) term. b) Find the sum of the first 5 terms. **Solution:** Here, \(a = 2\) and \(r = \frac{6}{2} = 3\). a) For the \(6^{th}\) term (\(n=6\)): \[ T_6 = ar^{6-1} = 2 \cdot 3^5 = 2 \cdot 243 = 486 \] The \(6^{th}\) term is 486. b) For the sum of the first 5 terms (\(n=5\)), since \(r=3 > 1\): \[ S_5 = \frac{a(r^5 - 1)}{r - 1} = \frac{2(3^5 - 1)}{3 - 1} = \frac{2(243 - 1)}{2} = 242 \] The sum of the first 5 terms is 242. #### Example 3: Finding the Geometric Mean Find the geometric mean between 5 and 20. **Solution:** Let \(a = 5\) and \(b = 20\). \[ G = \sqrt{ab} = \sqrt{5 \times 20} = \sqrt{100} = 10 \] The geometric mean between 5 and 20 is 10. ### Exercises 1. For the AP: 10, 13, 16, ... a) Find the \(15^{th}\) term. b) Find the sum of the first 20 terms. 2. For the GP: 1, 2, 4, 8, ... a) Find the \(8^{th}\) term. b) Find the sum of the first 7 terms. 3. The first term of an AP is 5 and the \(6^{th}\) term is 25. Find the common difference and the sum of the first 10 terms. 4. Find the geometric mean between: a) 3 and 27 b) 2 and 32 5. In a GP, the first term is 4 and the common ratio is \( -2 \). Find the sum of the first 6 terms. ### Assignment 1. An arithmetic sequence has \(T_3 = 18\) and \(T_7 = 30\). Find the first term, the common difference, and the sum of the first 15 terms. 2. The sum of the first 3 terms of a GP is 26 and the common ratio is 3. Find the first term and the sum of the first 5 terms. 3. A geometric sequence has a first term of 100 and a common ratio of \( \frac{1}{2} \). Which term is the first to be less than 1? 4. Insert two geometric means between 8 and 216. (Hint: if \(a, G_1, G_2, b\) form a GP, then \(b = ar^3\)). 5. A person saves 100 KES in the first month, 150 KES in the second month, 200 KES in the third month, and so on. How much money will they have saved after 2 years? ### Summary In this lesson, we explored the fascinating world of sequences, specifically Arithmetic Progressions (AP) and Geometric Progressions (GP). We learned that an AP has a **common difference** \(d\), with its \(n^{th}\) term given by \(T_n = a + (n-1)d\) and sum by \(S_n = \frac{n}{2}[2a + (n-1)d]\) or \(S_n = \frac{n}{2}(a+l)\). For GPs, we discovered a **common ratio** \(r\), with its \(n^{th}\) term as \(T_n = ar^{n-1}\) and sum as \(S_n = \frac{a(r^n-1)}{r-1}\) (or \(S_n = \frac{a(1-r^n)}{1-r}\)). Finally, we understood that the **Geometric Mean** (GM) between two positive numbers \(a\) and \(b\) is \(G = \sqrt{ab}\). Mastering these formulas and concepts will enable you to solve a wide range of problems involving growth, decay, and ordered lists of numbers. ## Sum of the first n terms of a GP and Application of ### Introduction Welcome to this lesson on Geometric Progressions (GPs)! In our previous lesson, we learned about what a Geometric Progression is, how to identify its first term \( a \) and common ratio \( r \), and how to find any specific term \( T_n \) using the formula \( T_n = ar^{n-1} \). Today, we will go a step further and learn how to find the sum of the first \( n \) terms of a GP. We will also explore the concept of the sum to infinity for certain GPs and look at various real-world applications of these concepts. ### Key Concepts #### Recall: Geometric Progression (GP) A Geometric Progression (GP), also known as a geometric sequence, is a sequence of non-zero numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio \( r \). * **First term:** \( a \) * **Common ratio:** \( r = \frac{T_2}{T_1} = \frac{T_3}{T_2} = \dots \) * **\( n \)th term:** \( T_n = ar^{n-1} \) #### Sum of the First \( n \) Terms of a GP (\( S_n \)) The sum of the first \( n \) terms of a GP is denoted by \( S_n \). There are two main cases to consider when deriving the formula for \( S_n \): ##### Case 1: When the common ratio \( r = 1 \) If \( r = 1 \), the GP looks like this: \( a, a(1), a(1)^2, ext{...}, a(1)^{n-1} \), which simplifies to \( a, a, a, ext{...}, a \). In this case, the sum of the first \( n \) terms is simply \( n \) times the first term. \[ S_n = a + a + a + ext{...} + a \quad (n \text{ times}) \] \[ S_n = na \] ##### Case 2: When the common ratio \( r \neq 1 \) Let the sum of the first \( n \) terms be: \[ S_n = a + ar + ar^2 + ext{...} + ar^{n-1} \quad \text{(Equation 1)} \] Multiply Equation 1 by \( r \): \[ rS_n = ar + ar^2 + ar^3 + ext{...} + ar^n \quad \text{(Equation 2)} \] Subtract Equation 2 from Equation 1: \[ S_n - rS_n = (a + ar + ext{...} + ar^{n-1}) - (ar + ar^2 + ext{...} + ar^n) \] \[ S_n(1 - r) = a - ar^n \] \[ S_n(1 - r) = a(1 - r^n) \] Therefore, if \( r \neq 1 \): \[ S_n = \frac{a(1 - r^n)}{1 - r} \] This formula is generally used when \( |r| < 1 \) (i.e., \( -1 < r < 1 \)). Alternatively, if we subtract Equation 1 from Equation 2, we get: \[ rS_n - S_n = (ar + ar^2 + ext{...} + ar^n) - (a + ar + ext{...} + ar^{n-1}) \] \[ S_n(r - 1) = ar^n - a \] \[ S_n(r - 1) = a(r^n - 1) \] Therefore, if \( r \neq 1 \): \[ S_n = \frac{a(r^n - 1)}{r - 1} \] This formula is generally used when \( |r| > 1 \) (i.e., \( r > 1 \) or \( r < -1 \)). Both formulas yield the same result, but choosing the appropriate one can help avoid negative denominators. #### Sum to Infinity of a GP (\( S_{\infty} \)) For a GP to have a sum to infinity, the terms of the sequence must get progressively smaller, approaching zero. This only happens when the common ratio \( r \) is between -1 and 1, exclusive. * **Condition for convergence:** \( |r| < 1 \) (i.e., \( -1 < r < 1 \)). If this condition is met, as \( n \) approaches infinity, \( r^n \) approaches 0. Using the formula \( S_n = \frac{a(1 - r^n)}{1 - r} \): \[ S_{\infty} = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{a(1 - r^n)}{1 - r} \] Since \( r^n \to 0 \) as \( n \to \infty \) when \( |r| < 1 \): \[ S_{\infty} = \frac{a(1 - 0)}{1 - r} \] \[ S_{\infty} = \frac{a}{1 - r} \] If \( |r| \geq 1 \), the terms either stay the same size or grow larger, so the sum to infinity does not exist (it diverges). #### Applications of Geometric Progressions GPs have numerous applications in real life, including: * **Compound Interest:** The growth of money with compound interest often follows a GP pattern. * **Depreciation:** The decrease in value of an asset over time at a fixed rate. * **Population Growth/Decay:** Modelling populations that grow or shrink by a certain percentage each period. * **Radioactive Decay:** The decrease in the amount of a radioactive substance over time. * **Rebounding Ball Problems:** The heights a ball reaches after successive bounces. ### Vocabulary * **Geometric Progression (GP):** A sequence where each term after the first is found by multiplying the previous one by a fixed non-zero number (common ratio). * **Common Ratio (\( r \)):** The constant factor between consecutive terms in a GP. * **Sum of the first \( n \) terms (\( S_n \)):** The sum of the first \( n \) terms of a GP. * **Sum to infinity (\( S_{\infty} \)):** The sum of an infinite number of terms in a GP, which exists only if \( |r| < 1 \). * **Convergent GP:** A GP for which the sum to infinity exists (i.e., \( |r| < 1 \)). * **Divergent GP:** A GP for which the sum to infinity does not exist (i.e., \( |r| \geq 1 \)). ### Examples **Example 1: Find the sum of the first 6 terms of the GP: 2, 6, 18, ...** Here, \( a = 2 \). The common ratio \( r = \frac{6}{2} = 3 \). Since \( r = 3 > 1 \), we use the formula \( S_n = \frac{a(r^n - 1)}{r - 1} \). We want to find \( S_6 \). \[ S_6 = \frac{2(3^6 - 1)}{3 - 1} \] \[ S_6 = \frac{2(729 - 1)}{2} \] \[ S_6 = \frac{2(728)}{2} \] \[ S_6 = 728 \] **Example 2: Find the sum of the first 5 terms of the GP: 16, 8, 4, ...** Here, \( a = 16 \). The common ratio \( r = \frac{8}{16} = \frac{1}{2} \). Since \( r = \frac{1}{2} < 1 \), we use the formula \( S_n = \frac{a(1 - r^n)}{1 - r} \). We want to find \( S_5 \). \[ S_5 = \frac{16(1 - (\frac{1}{2})^5)}{1 - \frac{1}{2}} \] \[ S_5 = \frac{16(1 - \frac{1}{32})}{\frac{1}{2}} \] \[ S_5 = \frac{16(\frac{31}{32})}{\frac{1}{2}} \] \[ S_5 = (16 \times \frac{31}{32}) \times 2 \] \[ S_5 = (\frac{31}{2}) \times 2 \] \[ S_5 = 31 \] **Example 3: Find the sum to infinity of the GP: 10, 5, 2.5, ...** Here, \( a = 10 \). The common ratio \( r = \frac{5}{10} = \frac{1}{2} \). Since \( |r| = |\frac{1}{2}| < 1 \), the sum to infinity exists. We use the formula \( S_{\infty} = \frac{a}{1 - r} \). \[ S_{\infty} = \frac{10}{1 - \frac{1}{2}} \] \[ S_{\infty} = \frac{10}{\frac{1}{2}} \] \[ S_{\infty} = 10 \times 2 \] \[ S_{\infty} = 20 \] **Example 4: Application - Population Growth** A town's population increases by 5% each year. If the current population is 20,000, what will be the total population over the next 4 years (including the current year)? This is a GP problem where: * First term \( a = 20,000 \) (current population). * Common ratio \( r = 1 + 0.05 = 1.05 \) (since it increases by 5%). * Number of years \( n = 4 \). We need to find the sum of populations for 4 years, starting from the current year's population. \[ S_4 = \frac{a(r^4 - 1)}{r - 1} \] \[ S_4 = \frac{20000((1.05)^4 - 1)}{1.05 - 1} \] \[ S_4 = \frac{20000(1.21550625 - 1)}{0.05} \] \[ S_4 = \frac{20000(0.21550625)}{0.05} \] \[ S_4 = 400000 \times 0.21550625 \] \[ S_4 = 86202.5 \] So, the total population over the next 4 years would be approximately 86,203 people. ### Exercises 1. Find the sum of the first 7 terms of the GP: 3, 9, 27, ... 2. Calculate \( S_5 \) for the GP: 64, 32, 16, ... 3. Determine the sum to infinity for the GP: 81, 27, 9, ... 4. A ball is dropped from a height of 10 meters. After each bounce, it rises to 80% of its previous height. Find the total vertical distance the ball travels until it comes to rest. 5. The first term of a GP is 5, and the common ratio is -2. Find the sum of the first 4 terms. ### Assignment 1. The sum of the first \( n \) terms of a GP is given by \( S_n = 3(2^n - 1) \). Find the first term and the common ratio of this GP. 2. In a GP, the first term is 12 and the sum to infinity is 18. Find the common ratio. 3. A company's sales are expected to increase by 10% each month. If the sales in the first month are $5000, what will be the total sales for the first 6 months? 4. The third term of a GP is 36, and the sixth term is 972. Find the sum of the first 5 terms of this GP. 5. Express the recurring decimal \( 0.\overline{36} \) as a fraction in its simplest form using the sum to infinity of a GP. ### Summary In this lesson, we have explored the concept of the sum of the first \( n \) terms of a Geometric Progression (GP) and its applications. We learned two main formulas for \( S_n \): * If \( r = 1 \), then \( S_n = na \). * If \( r \neq 1 \), then \( S_n = \frac{a(1 - r^n)}{1 - r} \) (often used for \( |r| < 1 \)) or \( S_n = \frac{a(r^n - 1)}{r - 1} \) (often used for \( |r| > 1 \)). We also discovered that for a GP where \( |r| < 1 \), the sum to infinity exists and is given by: \[ S_{\infty} = \frac{a}{1 - r} \] Finally, we looked at how these formulas can be applied to solve real-world problems involving compound interest, depreciation, and growth/decay models. Remember to always identify the first term \( a \), the common ratio \( r \), and the number of terms \( n \) (or check if \( |r| < 1 \) for infinite sums) before applying the correct formula. ## sequences to solve real life problems ### Introduction Welcome to this lesson on sequences! You might think sequences are just abstract numbers, but they are powerful mathematical tools used to solve many real-world problems. From calculating finances and population growth to understanding scientific phenomena, sequences help us model and predict outcomes. In this lesson, you will learn how to apply your knowledge of arithmetic and geometric sequences to solve practical problems encountered in everyday life. ### Key Concepts #### 1. Understanding Real-Life Problems as Sequences Many situations involve a pattern of numbers that increase or decrease in a predictable way. These patterns can often be represented by sequences. * **Identifying the Pattern:** The first step is to read the problem carefully and identify how the numbers are changing from one step to the next. * **Initial Value:** Determine the starting point or the first term of the sequence, denoted as \( a \). * **Change per Step:** Identify the constant change occurring at each step. This change will tell you whether the sequence is arithmetic or geometric. #### 2. Arithmetic Sequences in Real Life An arithmetic sequence is characterized by a constant difference between consecutive terms. This 'common difference' means quantities are increasing or decreasing by a fixed amount. ##### Characteristics of Arithmetic Problems: * Involve adding or subtracting a fixed quantity repeatedly. * Often describe linear growth or decay. * Examples: Simple interest, depreciation by a fixed amount, salary increases by a fixed sum each year, daily savings plans. ##### Key Formulas for Arithmetic Sequences: * **The \(n\)-th term:** \( T_n = a + (n-1)d \) * \( T_n \) = the value of the \(n\)-th term * \( a \) = the first term * \( n \) = the term number (e.g., 1st, 2nd, 3rd...) * \( d \) = the common difference * **Sum of the first \(n\) terms:** \[ S_n = \frac{n}{2} [2a + (n-1)d] \] or \[ S_n = \frac{n}{2} (a + T_n) \] * \( S_n \) = the sum of the first \(n\) terms #### 3. Geometric Sequences in Real Life A geometric sequence is characterized by a constant ratio between consecutive terms. This 'common ratio' means quantities are increasing or decreasing by a fixed percentage or factor. ##### Characteristics of Geometric Problems: * Involve multiplying or dividing by a fixed quantity repeatedly. * Often describe exponential growth or decay. * Examples: Compound interest, population growth/decline, radioactive decay, depreciation by a percentage, bacterial growth. ##### Key Formulas for Geometric Sequences: * **The \(n\)-th term:** \( T_n = ar^{n-1} \) * \( T_n \) = the value of the \(n\)-th term * \( a \) = the first term * \( n \) = the term number * \( r \) = the common ratio * **Sum of the first \(n\) terms:** \[ S_n = \frac{a(r^n - 1)}{r-1} \] for \( r \neq 1 \) * \( S_n \) = the sum of the first \(n\) terms #### 4. Problem-Solving Strategy 1. **Understand the Problem:** Read carefully to grasp what is being asked. 2. **Identify the Given Information:** Extract the initial value (\(a\)), the change per step (common difference \(d\) or common ratio \(r\)), and the number of terms (\(n\)). 3. **Determine the Type of Sequence:** * If a fixed amount is added or subtracted, it's an **arithmetic sequence**. * If a fixed amount (or percentage) is multiplied or divided, it's a **geometric sequence**. 4. **Choose the Correct Formula:** Select the appropriate formula for \( T_n \) or \( S_n \) based on the type of sequence and what needs to be calculated. 5. **Solve and Interpret:** Substitute the values into the formula and calculate the answer. Ensure your answer makes sense in the context of the problem. ### Vocabulary * **Sequence:** An ordered list of numbers. * **Term:** Each number in a sequence. * **Arithmetic Sequence:** A sequence where the difference between consecutive terms is constant. * **Common Difference (\(d\)):** The constant difference between consecutive terms in an arithmetic sequence. * **Geometric Sequence:** A sequence where the ratio of consecutive terms is constant. * **Common Ratio (\(r\)):** The constant ratio between consecutive terms in a geometric sequence. * **\(n\)-th Term (\(T_n\)):** The value of the term at a specific position \(n\) in the sequence. * **Sum of \(n\) Terms (\(S_n\)):** The total sum of the first \(n\) terms in a sequence. ### Examples #### Example 1: (Arithmetic Sequence - Savings) A student saves \(RM 200\) in the first month. In each subsequent month, they save an additional \(RM 50\) more than the previous month. How much money will they have saved in total after 12 months? **Solution:** 1. **Identify:** This is an arithmetic sequence because a fixed amount (\(RM 50\)) is added each month. 2. **Given:** * First term, \(a = 200\) * Common difference, \(d = 50\) * Number of months, \(n = 12\) 3. **Required:** Total savings after 12 months, which is the sum of the first 12 terms (\(S_{12}\)). 4. **Formula:** We use the sum of an arithmetic sequence formula: \[ S_n = \frac{n}{2} [2a + (n-1)d] \] 5. **Calculate:** \[ S_{12} = \frac{12}{2} [2(200) + (12-1)50] \] \[ S_{12} = 6 [400 + (11)50] \] \[ S_{12} = 6 [400 + 550] \] \[ S_{12} = 6 [950] \] \[ S_{12} = 5700 \] **Answer:** The student will have saved \(RM 5700\) in total after 12 months. #### Example 2: (Geometric Sequence - Population Growth) The population of a town is 10,000. If the population increases by 3% each year, what will the population be after 5 years? (Give your answer to the nearest whole number). **Solution:** 1. **Identify:** This is a geometric sequence because the population increases by a percentage (multiplicative factor) each year. 2. **Given:** * First term (initial population), \(a = 10000\) * Rate of increase = 3% = 0.03 * Common ratio, \(r = 1 + 0.03 = 1.03\) (since it increases by 3%) * Number of years, \(n = 5\) (We are looking for the population *after* 5 years, which means the 6th term if year 0 is the 1st term, or the 5th *increase* from the initial value. Let's use \(n=6\) for \(T_6\) if \(a\) is year 0, or calculate after 5 increases from the initial \(a\) using \(T_{n+1}\). More directly, after 5 years, it's \(a imes r^5\) if \(a\) is initial, so we're looking for the value at the end of the 5th year, meaning \(ar^n\) or \(T_{n+1}\). Let's use \(n=5\) in \(ar^n\) for the final value after \(n\) periods, or \(T_{n+1} = ar^n\) if \(a\) is the starting point (Year 0, term 1). Using \(T_n = ar^{n-1}\) means for 5 years *of growth*, we are looking for the 6th term (initial + 5 increases), so \(n=6\). Let's rephrase: if \(a\) is population at start of year 1 (Year 0), then after 1 year it's \(ar^1\), after 2 years it's \(ar^2\), so after 5 years it's \(ar^5\). This means we use \(a = 10000\), \(r = 1.03\) and raise \(r\) to the power of \(5\) to find the population *after* 5 years. This is equivalent to \(T_{n+1} = ar^n\) where \(n\) is the number of increases. 3. **Required:** Population after 5 years (which is the result of 5 growth periods). 4. **Formula:** We use a modified geometric sequence formula: \( P_n = a r^n \) where \(n\) is the number of periods/years, and \(P_n\) is the value *after* \(n\) periods. 5. **Calculate:** \[ P_5 = 10000 (1.03)^5 \] \[ P_5 = 10000 (1.1592740743) \] \[ P_5 = 11592.740743 \] Rounding to the nearest whole number: \[ P_5 \approx 11593 \] **Answer:** The population of the town will be approximately 11,593 after 5 years. ### Exercises 1. **Exercise 1 (Arithmetic):** A cyclist trains by cycling \(30 ext{ km}\) in the first week. Each week, she increases her distance by \(5 ext{ km}\). How far will she cycle in the 10th week? What is the total distance she cycles in the first 10 weeks? 2. **Exercise 2 (Geometric):** A bacterial culture starts with 500 bacteria. The number of bacteria doubles every hour. How many bacteria will there be after 6 hours? 3. **Exercise 3 (Mixed):** A car bought for \(RM 80,000\) depreciates by \(15\%\) per year. Calculate its value after 3 years. ### Assignment 1. **Assignment 1:** A concert hall has 25 seats in the first row, 28 seats in the second row, 31 seats in the third row, and so on. If there are 20 rows in total, how many seats are there in the last row? What is the total seating capacity of the hall? 2. **Assignment 2:** A certain radioactive substance decays such that its mass decreases by \(10\%\) every year. If the initial mass is \(200 ext{ grams}\), what mass of the substance will remain after 7 years? (Round your answer to two decimal places). 3. **Assignment 3:** A new company's profits increased by \(RM 15,000\) in its first year, then by \(RM 17,000\) in its second year, \(RM 19,000\) in its third year, and so on. Assuming this pattern continues, what will be the company's profit increase in the 8th year? What is the total profit increase over the first 8 years? ### Summary In this lesson, you've learned how to apply your knowledge of arithmetic and geometric sequences to solve real-life problems. The key is to first identify whether the problem involves a constant difference (arithmetic) or a constant ratio (geometric). Once identified, you can use the appropriate formulas: * **Arithmetic Sequence:** \( T_n = a + (n-1)d \) and \[ S_n = \frac{n}{2} [2a + (n-1)d] \] * **Geometric Sequence:** \( T_n = ar^{n-1} \) and \[ S_n = \frac{a(r^n - 1)}{r-1} \] By following a systematic problem-solving approach, you can model various real-world scenarios, from financial planning to population changes, using the power of sequences.