Problem 26 - College Algebra

Question

Use the following scenario. Javier makes monthly deposits into a savings account. He opened the account with an initial deposit of \(\$ 50 .\) Each month thereafter he increased the previous deposit amount by \(\$ 20\). Graph the arithmetic sequence showing one year of Javier's deposits.

Step-by-Step Solution

Answer

The deposits form an arithmetic sequence starting at \(50\) in month 1 and increasing by \(20\) each month, ending at \(270\) in month 12.

Step 1: Identify the Sequence

The sequence of Javier's deposits can be represented as an arithmetic sequence because the difference between successive deposits is constant. The initial deposit (first term) is \( a_1 = 50 \), and the common difference \( d = 20 \).

Step 2: Determine the Sequence Terms

To find the terms of the sequence for the first 12 months, we calculate each term using the formula for an arithmetic sequence. The general formula is \( a_n = a_1 + (n-1) \cdot d \), where \( n \) is the month number. For example, the deposit in the 2nd month is \( a_2 = 50 + (2-1) \cdot 20 = 70 \).

Step 3: Calculate All 12 Terms

Use the formula to calculate the first 12 deposits:- Month 1: \( a_1 = 50 \)- Month 2: \( a_2 = 70 \)- Month 3: \( a_3 = 90 \)- Month 4: \( a_4 = 110 \)- Month 5: \( a_5 = 130 \)- Month 6: \( a_6 = 150 \)- Month 7: \( a_7 = 170 \)- Month 8: \( a_8 = 190 \)- Month 9: \( a_9 = 210 \)- Month 10: \( a_{10} = 230 \)- Month 11: \( a_{11} = 250 \)- Month 12: \( a_{12} = 270 \)

Step 4: Graph the Sequence

Plot the points on a graph with the x-axis representing the months (1 through 12) and the y-axis representing the deposit amounts. Connect the points to form a line, as the sequence is arithmetic and therefore linear.

Step 5: Analyze the Graph

The graph will show a straight line as arithmetic sequences have a constant rate of increase. The line starts at \( \(1, 50\) \) and ends at \( \(12, 270\) \). This demonstrates how Javier's deposits increase by \( 20 \) each month.

Key Concepts

Graphing Arithmetic Sequences

Graphing an arithmetic sequence is a straightforward task once you understand the concepts behind the sequence. In the case of Javier's deposits, you have an arithmetic sequence because there is a constant difference between consecutive terms. Each term represents the amount Javier deposits each month.
To graph the sequence:

• Plot each deposit amount along the y-axis.
• Use the months as values on the x-axis.
• Each point represents the deposit amount Javier made every month.
• Connect these points with a straight line.

The line should rise steadily upwards, showing a linear relationship because each month the deposit increases by the same amount. This linear graph visually illustrates the consistent, regular increase in Javier's savings, making it easy to see the arithmetic nature of the deposits.

Common Difference

The common difference in an arithmetic sequence is a crucial element that sets the pattern of the sequence. In simpler terms, it's the amount that the sequence increases or decreases by every time.
In Javier's scenario:

• The initial deposit is \(50.
• Every month, he increases his deposit by \)20.

This \(20 is the common difference, represented as \(d = 20\).
Understanding the common difference allows anyone to predict future terms of the sequence effortlessly. You know that each term will simply be the previous term plus \)20 in this case. This consistency means you can always find future deposit amounts without starting over from the first term.

Arithmetic Sequence Formula

The arithmetic sequence formula is a valuable tool for calculating the terms of a sequence without listing them all out. For any arithmetic sequence, you can find the nth term using:\[a_n = a_1 + (n-1) \cdot d\]where:

• \(a_n\) is the nth term you want to find.
• \(a_1\) is the first term of the sequence. For Javier, that's \(50.
• \(n\) is the term number, representing the month in this example.
• \(d\) is the common difference, \)20 for Javier's deposits.

This formula is like a shortcut for quickly finding any term. If you want to know the deposit amount for the 7th month, replace n with 7, and you'll compute the amount as:\[a_7 = 50 + (7-1) \cdot 20 = 170\]Using this formula, you can easily analyze sequences without cumbersome calculations for every month separately.