Problem 21 - College Algebra with Corequisite Support

Question

For the following exercises, find the first term given two terms from an arithmetic sequence. Find the first term or \(a_{1}\) of an arithmetic sequence if \(a_{8}=40\) and \(a_{23}=115\).

Step-by-Step Solution

Answer

The first term \( a_1 \) is 5.

Step 1: Understand the formula for an arithmetic sequence

In an arithmetic sequence, the nth term is given by the formula \( a_n = a_1 + (n-1) \cdot d \), where \( a_n \) is the nth term, \( a_1 \) is the first term, and \( d \) is the common difference between terms.

Step 2: Set up equations for given terms

We have \( a_8 = 40 \) and \( a_{23} = 115 \). Using the formula from Step 1, we can set up the equations: \( a_8 = a_1 + 7d = 40 \) and \( a_{23} = a_1 + 22d = 115 \).

Step 3: Solve the system of equations

Subtract the first equation from the second to eliminate \( a_1 \):\[(a_1 + 22d) - (a_1 + 7d) = 115 - 40 \15d = 75\]Divide by 15 to find \( d \): \( d = 5 \).

Step 4: Substitute to find the first term

Use the value of \( d \) found in Step 3 in one of the original equations to solve for \( a_1 \). Using \( a_8 = a_1 + 7 \cdot 5 = 40 \), solve for \( a_1 \):\[a_1 + 35 = 40 \a_1 = 40 - 35 \a_1 = 5\]

Step 5: Verify the solution

To ensure the solution is correct, substitute \( a_1 = 5 \) and \( d = 5 \) back into the equation for \( a_{23} \): \[a_{23} = 5 + 22 \cdot 5 = 5 + 110 = 115\]Both conditions are satisfied, confirming \( a_1 = 5 \) is correct.

Key Concepts

Common Difference

In an arithmetic sequence, the **common difference** is a crucial element. It refers to the consistent amount by which each term in the sequence increases (or decreases) from the previous term. One of its key properties is that it remains constant throughout the sequence.

The formula defining the sequence is:

• \( a_n = a_1 + (n-1) \cdot d \)

Here, \( d \) is the common difference. In practical terms, it's what you add to get from one term to the next. Knowing \( d \), you can easily find any term if you know the first term.

To calculate \( d \), one can simply subtract one term from the next. If given more than one term, like in the exercise, you can set up an equation involving known terms and solve for \( d \) as demonstrated in the original step-by-step solution. This provides a foundation for unraveling the entire sequence including calculating any specific term.

First Term Calculation

When faced with an arithmetic sequence, finding the **first term**, \( a_1 \), is often the initial task that paves the way to understand the sequence as a whole. It's the starting point from which all other terms are derived using the common difference.

To determine \( a_1 \), we connect a known term to the sequence formula:

• \( a_n = a_1 + (n-1) \cdot d \)

Rearranging this to solve for \( a_1 \):

• \( a_1 = a_n - (n-1) \cdot d \)

This equation shows us how to step "back" to \( a_1 \) using any term \( a_n \) provided the common difference \( d \) is known.

In our specific exercise, from knowing \( a_8 \) and \( a_{23} \), we ultimately used the formula to reach back from \( a_8 \) to \( a_1 \), leveraging the value of \( d \) obtained from solving the system of equations. This straightforward yet powerful calculation method shows how critical the first term is in framing an arithmetic sequence.

System of Equations

Using a **system of equations** is a robust strategy for solving problems in arithmetic sequences when given multiple terms. The idea is to set up equations that represent each known term as a point in the sequence.

In the original problem, we were given two equations:

• \( a_8 = a_1 + 7 \cdot d = 40 \)
• \( a_{23} = a_1 + 22 \cdot d = 115 \)

By subtracting one equation from another, we eliminate \( a_1 \) and solve for the common difference \( d \). This approach simplifies complex dependencies and allows us to break down the problem into smaller, manageable parts.

The value of \( d \) found through this method can then be substituted back into one of the original equations to find \( a_1 \).

This technique not only verifies the consistency of the given terms but also solidifies our grasp of the sequence's structure. Solving these equations showcases the utility of basic algebra in extracting significant insights in arithmetic sequences.