Arithmetic Sequence Word Problems Worksheet

Mastering Arithmetic Sequence Word Problems: A Comprehensive Guide

This worksheet tackles arithmetic sequences, a fundamental concept in algebra. Understanding arithmetic sequences involves recognizing patterns, applying formulas, and solving real-world problems. This guide will walk you through various types of word problems, providing step-by-step solutions and strategies to master this important topic. We will cover everything from basic problems to more challenging scenarios, equipping you with the skills to confidently tackle any arithmetic sequence word problem.

Understanding Arithmetic Sequences

Before diving into word problems, let's review the core concepts. An arithmetic sequence is a list of numbers where the difference between consecutive terms remains constant. This constant difference is called the common difference, often denoted as 'd'.

Key Elements:

• First term (a₁): The first number in the sequence.
• Common difference (d): The constant difference between consecutive terms.
• nth term (aₙ): The term at the nth position in the sequence.

Formulas:

The two most important formulas for working with arithmetic sequences are:

• Finding the nth term: aₙ = a₁ + (n-1)d
• Finding the sum of the first n terms: Sₙ = n/2 [2a₁ + (n-1)d] or Sₙ = n/2 (a₁ + aₙ)

Types of Arithmetic Sequence Word Problems and Solved Examples

Let's explore different types of word problems involving arithmetic sequences and how to solve them systematically.

Type 1: Finding the nth term

Problem 1: A theater has 25 seats in the first row. Each subsequent row has 4 more seats than the previous row. How many seats are there in the 10th row?

Solution:

1. Identify the key elements:

   • a₁ (first term) = 25 seats
   • d (common difference) = 4 seats
   • n (number of terms) = 10

2. Apply the formula for the nth term: aₙ = a₁ + (n-1)d

3. Substitute the values: a₁₀ = 25 + (10-1)4 = 25 + 36 = 61 seats

Answer: There are 61 seats in the 10th row.

Problem 2: A plant grows 2cm taller each day. If it starts at a height of 5cm, what will its height be after 15 days?

Solution:

1. Identify the key elements:

   • a₁ = 5 cm
   • d = 2 cm
   • n = 15 days

2. Apply the formula: aₙ = a₁ + (n-1)d

3. Substitute the values: a₁₅ = 5 + (15-1)2 = 5 + 28 = 33 cm

Answer: The plant will be 33 cm tall after 15 days.

Type 2: Finding the common difference

Problem 3: The 5th term of an arithmetic sequence is 22 and the 10th term is 47. Find the common difference.

Solution:

1. Use the formula for the nth term: We have two equations:

   • a₅ = a₁ + 4d = 22
   • a₁₀ = a₁ + 9d = 47

2. Solve the system of equations: Subtract the first equation from the second:

   • (a₁ + 9d) - (a₁ + 4d) = 47 - 22
   • 5d = 25
   • d = 5

Answer: The common difference is 5.

Type 3: Finding the number of terms

Problem 4: A stack of logs has 20 logs in the bottom row, 18 in the second row, and so on, decreasing by 2 logs each row. If the top row has 2 logs, how many rows are there?

Solution:

1. Identify the key elements:

   • a₁ = 20
   • d = -2
   • aₙ = 2

2. Use the formula for the nth term: aₙ = a₁ + (n-1)d

3. Substitute the values and solve for n: 2 = 20 + (n-1)(-2)

   • -18 = (n-1)(-2)
   • 9 = n-1
   • n = 10

Answer: There are 10 rows of logs.

Type 4: Finding the sum of an arithmetic sequence

Problem 5: A person saves $10 in the first week, $13 in the second week, $16 in the third week, and so on. How much money will they have saved after 20 weeks?

Solution:

1. Identify the key elements:

   • a₁ = 10
   • d = 3
   • n = 20

2. Use the formula for the sum of an arithmetic series: Sₙ = n/2 [2a₁ + (n-1)d]

3. Substitute the values: S₂₀ = 20/2 [2(10) + (20-1)3] = 10[20 + 57] = 770

Answer: The person will have saved $770 after 20 weeks.

Type 5: Real-world Applications involving patterns and sequences.

Problem 6: A bricklayer is building a wall. The first layer has 10 bricks. Each subsequent layer has 2 more bricks than the layer below. How many bricks are needed to build a wall with 12 layers?

Solution: This problem combines finding the number of bricks in the 12th layer and then summing the bricks across all layers.

1. Find the number of bricks in the 12th layer:

   • a₁ = 10
   • d = 2
   • n = 12
   • a₁₂ = a₁ + (n-1)d = 10 + (12-1)2 = 38 bricks

2. Find the total number of bricks: We need the sum of the arithmetic series.

   • Sₙ = n/2 (a₁ + aₙ) = 12/2 (10 + 38) = 6(48) = 288 bricks

Answer: 288 bricks are needed to build the wall.

Advanced Arithmetic Sequence Word Problems

Let's tackle some more complex scenarios requiring a deeper understanding of the concepts.

Problem 7: The sum of the first 8 terms of an arithmetic sequence is 100, and the 8th term is 22. Find the first term and the common difference.

Solution:

1. Use the sum formula: S₈ = 8/2 (a₁ + a₈) = 100. This simplifies to 4(a₁ + 22) = 100.

2. Solve for a₁: a₁ + 22 = 25, a₁ = 3.

3. Use the nth term formula: a₈ = a₁ + 7d = 22. Substitute a₁ = 3: 3 + 7d = 22.

4. Solve for d: 7d = 19, d = 19/7.

Answer: The first term (a₁) is 3, and the common difference (d) is 19/7.

Problem 8: A ball is dropped from a height of 10 meters. Each time it bounces, it reaches 80% of its previous height. Model this situation using an arithmetic or geometric sequence, and find the total distance the ball travels after 5 bounces. Note that this requires the use of a geometric sequence (not an arithmetic one)

Solution: This is not an arithmetic sequence, as the height decreases by a percentage, not a constant value. This is a geometric sequence. However, given the context of this worksheet, it's important to understand why it's not an arithmetic sequence. Arithmetic sequences have a constant difference, while geometric sequences have a constant ratio. We will still calculate the total distance traveled.

1. Identify the elements of the geometric sequence:

   • First term (a₁) = 10 meters (initial height)
   • Common ratio (r) = 0.8 (80% of the previous height)

2. Calculate the height after each bounce: This will create the geometric sequence.

3. Calculate the total distance: The total distance is the sum of the initial drop and the distance traveled during each bounce. This needs a geometric series summation. The formula is: S = a₁(1-rⁿ)/(1-r) where 'S' is the sum, 'a₁' is the first term, 'r' is the common ratio and 'n' is the number of terms. In this case, we need to double the sum (excluding the initial drop) because it bounces up and down.

   • Let's calculate the total distance, considering both up and down: 10 + 210(0.8) + 210(0.8)² + 210(0.8)³ + 210(0.8)⁴ = 10 + 16 + 12.8 + 10.24 + 8.192 = 57.232 meters

Answer: The ball travels approximately 57.23 meters after 5 bounces.

Frequently Asked Questions (FAQ)

Q1: What is the difference between an arithmetic sequence and a geometric sequence?

A: An arithmetic sequence has a constant difference between consecutive terms, while a geometric sequence has a constant ratio between consecutive terms.

Q2: How can I identify if a word problem involves an arithmetic sequence?

A: Look for keywords like "increases by," "decreases by," "adds," "subtracts," or situations where a constant amount is added or subtracted repeatedly.

Q3: What if I don't know the first term or the common difference?

A: You can often use a system of equations to solve for the missing values using the given information, as shown in Problem 7.

Conclusion

Mastering arithmetic sequence word problems requires a solid understanding of the fundamental concepts, formulas, and problem-solving strategies. By practicing a variety of problems, from basic to advanced, you'll build your confidence and develop the skills to tackle any challenge. Remember to always carefully identify the key elements (first term, common difference, number of terms) and choose the appropriate formula to solve the problem efficiently. Consistent practice is key to success! Remember to always check your work and ensure your answer makes logical sense within the context of the problem. Good luck!