Problem 59 - College Algebra

Question

Use a system of two equations in two variables, \(a_{1}\) and \(d,\) to solve Exercises \(59-60\) Write a formula for the general term (the \(n\) th term) of the arithmetic sequence whose second term, \(a_{2},\) is 4 and whose sixth term, \(a_{6}\) is 16

Step-by-Step Solution

Answer

The formula for the nth term, \(a_n\), in the arithmetic sequence is \(a_n = 1 + (n-1)*3\)

Step 1: Write down the given equations

We know that the nth term of an arithmetic sequence can be written as \(a_n = a_1 + (n-1)*d\). From the question, we understand that \(a_2 = 4\) and \(a_6 = 16\). Substituting these into our formula we get two equations: 1. \(4 = a_1 + d\) 2. \(16 = a_1 + 5d\)

Step 2: Solve the system of equations

Now we have a system of two equations and two variables that we can solve. One way of doing this is by subtracting equation 1 from equation 2:\(16 - 4 = (a_1 + 5d) - (a_1 + d)\)This simplifies to:\(12 = 4d\)Solving for \(d\) we get \(d = 3\)We can then substitute \(d\)=3 into equation 1 to solve for \(a_1\):\(4 = a_1 + 3\)So, \(a_1 = 1\)

Step 3: Write down the formula for the nth term

Now we have found \(a_1\) and \(d\), we can substitute these values back into the formula for the nth term of an arithmetic sequence. Thus the formula is \(a_n = 1 + (n-1)*3\)

Key Concepts

System of Equations

In mathematics, a system of equations is a set of two or more equations that involve the same set of variables. These equations are linked together; solving the system means finding the values of the variables that satisfy all the equations simultaneously. Systems are used in various fields like physics, economics, and engineering to describe a set of conditions that occur together.

For example, in the exercise provided, the system consists of two linear equations. We set them up based on the information given about an arithmetic sequence. The equations are linked through the common variables: the first term, denoted as \(a_1\), and the common difference, denoted as \(d\). By solving this system, we determine the specific arithmetic sequence that meets the provided conditions.

Arithmetic Sequence

An arithmetic sequence is a sequence of numbers in which the difference between consecutive terms is constant. This difference is known as the 'common difference' and is denoted by the variable \(d\). Arithmetic sequences are a foundational concept in algebra and are governed by a specific formula.

An important characteristic of an arithmetic sequence is its predictability; once you know the first term and the common difference, you can easily find any term in the sequence. The sequence progresses by adding the common difference to the previous term repeatedly. In our exercise, knowing the second and sixth terms allowed for the deduction of both the first term and the common difference.

Nth Term of a Sequence

The 'nth term' represents any generic term in a sequence. It's the formula that provides the value of the term at any given position \(n\) within the sequence. For an arithmetic sequence, the nth term is calculated using the following formula: \[a_n = a_1 + (n-1)d\]

Here, \(a_n\) represents the nth term of the sequence, \(a_1\) is the first term, and \(d\) is the common difference. By manipulating this formula with given terms, as was done in the exercise, you can find unknown variables and then express the nth term in a manner that allows calculation of any term based on its position in the sequence.

Solving Linear Equations

Linear equations are equations between two variables that produce a straight line when graphed. They have one or more terms that are either constants or the product of a constant and a single variable. Solving linear equations is about finding the values for the variables that make the equation true.

In the given exercise, we solved a linear system by substitution and elimination methods. First, one of the equations was reorganized to express one variable in terms of the other. Then, this expression was used to replace the variable in the second equation. Simplification led to the discovery of one variable's value, which was then used to find the other's, illustrating a classic approach to solving linear equations in a system.