Problem 50 - College Algebra with Modeling and Visualization

Question

Find a general term \(a_{n}\) for the arithmetic sequence. $$a_{3}=10, a_{7}=-4$$

Step-by-Step Solution

Answer

The general term is \(a_n = 20.5 - 3.5n\).

Step 1: Identify the general term of an arithmetic sequence

The general term of an arithmetic sequence is given by \( a_n = a_1 + (n-1)d \), where \( a_1 \) is the first term and \( d \) is the common difference between successive terms.

Step 2: Set up equations using given terms

Substitute the given terms into the general term formula to create equations. For the third term, \( a_3 = a_1 + 2d = 10 \), and for the seventh term, \( a_7 = a_1 + 6d = -4 \).

Step 3: Solve the system of equations

Subtract the first equation from the second to eliminate \( a_1 \) and find \( d \): \[ (a_1 + 6d) - (a_1 + 2d) = -4 - 10 \] This simplifies to \( 4d = -14 \), giving \( d = -3.5 \).

Step 4: Find the first term \(a_1\)

Substitute \( d = -3.5 \) back into one of the original equations, say \( a_1 + 2d = 10 \), to find \( a_1 \): \( a_1 + 2(-3.5) = 10 \), which simplifies to \( a_1 - 7 = 10 \). So, \( a_1 = 17 \).

Step 5: Write the general term formula

Substitute \( a_1 = 17 \) and \( d = -3.5 \) into the general term formula: \( a_n = 17 + (n-1)(-3.5) \). Simplify this to get the final form: \( a_n = 17 - 3.5n + 3.5 \), which is \( a_n = 20.5 - 3.5n \).

Key Concepts

General Term Formula

The general term formula is like the secret recipe for any arithmetic sequence. It helps to find any term in the sequence if you know the first term and the common difference. The formula is expressed as:
\( a_n = a_1 + (n-1)d \)
Here, \( a_n \) denotes the \( n^{th} \) term, \( a_1 \) is the first term, and \( d \) represents the common difference between terms in the sequence.
This formula is powerful because it can give you the value of any term, no matter how large \( n \) is, by simply plugging the known values into it.
Remember the two golden items needed: the first term \( a_1 \) and the common difference \( d \). Let's now see how the common difference plays a critical role.

Common Difference

The common difference \( d \) is key to understanding arithmetic sequences. It tells you how much each term in the sequence is increasing or decreasing by as you move from one term to the next.
Here's how you can conceptualize it:

• If \( d \) is positive, the sequence grows larger with each step.
• If \( d \) is negative, each term is smaller than the previous term.
• If \( d \) is zero, all terms are the same, as there is no change from term to term.

In the context of our problem, by creating equations with the given terms of the sequence, we deductively found \( d = -3.5 \). This means the sequence diminishes by 3.5 units as each new term is generated.
The common difference not only links terms together but also takes you a step closer to modeling the entire sequence and applying it to solve problems.

System of Equations

When given specific terms of an arithmetic sequence, establishing a system of equations is like pulling out the building blocks of the sequence.
In our example, with \( a_3 = 10 \) and \( a_7 = -4 \), we derived two separate equations using the general term formula:
\[ a_3 = a_1 + 2d = 10 \]\[ a_7 = a_1 + 6d = -4 \]This system of equations allows us to unravel unknowns, specifically \( a_1 \) and \( d \).
To solve, we can subtract the first equation from the second, thus eliminating \( a_1 \) and directly solving for the common difference \( d \). This showcases the strategic elimination approach often used in algebra to simplify and solve for multiple variables.

Solve for First Term

Once you identify the common difference, solving for the first term \( a_1 \) becomes straightforward.
You can substitute the value of \( d \) back into any of the original equations. For example, we used:\[ a_1 + 2(-3.5) = 10 \]This simplified to:\[ a_1 - 7 = 10 \]Hence, \( a_1 = 17 \).
This calculated \( a_1 \) completes our understanding of the sequence and empowers us to apply the general term formula for any \( n \). Recalling that this step is often the concluding part of solving sequence problems, it reassures that our found \( a_1 \) aligns perfectly with our earlier findings, allowing confidence in the sequence's representation.