Problem 20
Question
On July \(1,\) Mr. Taylor owed \(\$ 6,000 .\) On the 1 st of each of the following months, he repaid \(\$ 400\) . List the amount owed by Mr. Taylor on the 2nd of each month starting with July 2 . Explain why the amount owed each month forms an arithmetic sequence.
Step-by-Step Solution
Verified Answer
Mr. Taylor's debt on the 2nd of each month is: $6000, $5600, $5200, $4800, $4400. This forms an arithmetic sequence with a common difference of $400.
1Step 1: Initial Amount Owed
On July 1, Mr. Taylor owed $6000. We need to start with this amount to calculate the subsequent months' dues.
2Step 2: Determine Monthly Payment
Each month, Mr. Taylor repays $400. This means that every subsequent month, the amount he owes decreases by $400.
3Step 3: Calculate Amount Owed on July 2
Since no payment has been made yet by July 2, the amount owed remains at $6000.
4Step 4: Calculate Amount Owed on August 2
By August 2, Mr. Taylor has made one repayment of $400, so:\[ 6000 - 400 = 5600 \]
5Step 5: Calculate Amount Owed on September 2
By September 2, Mr. Taylor has made two repayments, reducing the debt by $800 total:\[ 6000 - 2(400) = 6000 - 800 = 5200 \]
6Step 6: Calculate Amount Owed on October 2
On October 2, Mr. Taylor has repaid for three months, amounting to $1200:\[ 6000 - 3(400) = 6000 - 1200 = 4800 \]
7Step 7: Calculate Amount Owed on November 2
By November 2, Mr. Taylor has repaid $1600 in total over four months:\[ 6000 - 4(400) = 6000 - 1600 = 4400 \]
8Step 8: Identify Arithmetic Sequence Pattern
The amount owed each month (6000, 5600, 5200, 4800, 4400) forms an arithmetic sequence because each subsequent term decreases by a constant amount of $400. In an arithmetic sequence, the difference between consecutive terms is constant, which is evident here as the common difference is $400.
Key Concepts
Arithmetic SeriesAlgebraMonthly Repayment Schedule
Arithmetic Series
An arithmetic series is a sequence of numbers in which the difference between consecutive terms is constant. This is called the common difference. In the case of Mr. Taylor's repayment schedule, the amount he owed each month forms an arithmetic sequence. Initially, he owed $6000, and by repaying $400 each month, the amount he owed decreased steadily.
To understand this better, you can think of an arithmetic sequence like a set of evenly spaced numbers on a number line. For example:
To understand this better, you can think of an arithmetic sequence like a set of evenly spaced numbers on a number line. For example:
- 6000 (initial amount owed)
- 5600 (owed after one payment)
- 5200 (owed after two payments)
- 4800 (owed after three payments)
- 4400 (owed after four payments)
Algebra
Algebra is a branch of mathematics that uses symbols and letters to represent numbers and quantities in formulas and equations. It is helpful in forming expressions and creating equations that solve real-world problems, such as determining the amount owed over time in a repayment plan.
Let's see how algebra comes into play with Mr. Taylor's debt:
Let's see how algebra comes into play with Mr. Taylor's debt:
- Let the initial amount owed be represented by \( A \).
- The monthly repayment is \( d = 400 \).
- For any month, \( n \), the amount owed \( A_n \) can be calculated as \( A - n \times d \).
Monthly Repayment Schedule
A monthly repayment schedule is a plan that outlines how much a borrower will pay each month to settle a debt. It helps both the lender and the borrower keep track of the loan's status and manage finances effectively.
For Mr. Taylor's repayment schedule, it's crucial to understand:
For Mr. Taylor's repayment schedule, it's crucial to understand:
- How much should be paid each month (here it's $400).
- The starting amount owed (in this case, $6000).
- The duration of repayments (each month calculates how much is left).
Other exercises in this chapter
Problem 20
In \(19-30 :\) a. Write an algebraic expression that represents \(a_{n}\) for each sequence. b. Find the ninth term of each sequence. $$ 3,6,9,12, \dots $$
View solution Problem 20
In \(19-24 :\) a. Write each arithmetic series as the sum of terms. b. Find the sum. $$ \sum_{k=2}^{8}(3+k) $$
View solution Problem 21
In \(15-26,\) write each series in sigma notation. $$ \frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+\frac{5}{6}+\frac{6}{7}+\frac{7}{8}+\frac{8}{9}+\frac{9}{1
View solution Problem 21
In \(15-22 :\) a. Write each sum as a series. b. Find the sum of each series. $$ 100+\sum_{k=1}^{6} 100\left(\frac{1}{2}\right)^{k} $$
View solution