Problem 32 - Algebra for College Students

Question

If the 5 th term of an arithmetic sequence is 14 and the 12 th term is 42 , find the first term.

Step-by-Step Solution

Answer

The first term is -2.

Step 1: Understanding the Problem

An arithmetic sequence is defined as a sequence of numbers in which the difference between any two consecutive terms is constant, called the common difference \(d\). Given the 5th term and the 12th term, we need to find the first term \(a_1\) of the sequence.

Step 2: Formula for the n-th Term

The formula for the n-th term \(a_n\) of an arithmetic sequence is given by: \(a_n = a_1 + (n-1) \, d\). We can use this formula to write equations for the given terms.

Step 3: Set Up Equations for Given Terms

Using the formula, the equation for the 5th term is: \(a_5 = a_1 + 4d = 14\). The equation for the 12th term is: \(a_{12} = a_1 + 11d = 42\).

Step 4: Subtraction to Eliminate \(a_1\)

Subtract the first equation from the second equation to eliminate \(a_1\):\((a_1 + 11d) - (a_1 + 4d) = 42 - 14\).This simplifies to: \(7d = 28\).

Step 5: Solve for the Common Difference \(d\)

Solving the equation \(7d = 28\), we find:\[d = \frac{28}{7} = 4\].Thus, the common difference \(d\) is 4.

Step 6: Substitute \(d\) Back to Find \(a_1\)

Using \(d=4\) in the equation \(a_1 + 4d = 14\), we substitute:\(a_1 + 4 \times 4 = 14\) which simplifies to \(a_1 + 16 = 14\).Solving for \(a_1\) gives: \(a_1 = 14 - 16 = -2\).

Key Concepts

Common Difference in Arithmetic Sequences

In an arithmetic sequence, each term after the first is generated by adding a constant called the "common difference" to the previous term. This constant is what defines the sequence as arithmetic.
Understanding the common difference, denoted as \(d\), is crucial when working with arithmetic sequences. When the difference between each term remains consistent, it allows for predictions and calculations of other terms.
For example, if you know one term in the sequence and the common difference, you can find subsequent terms by simple addition or subtraction.

• The common difference \(d\) can be positive, negative, or zero. A positive \(d\) means the sequence is increasing, a negative \(d\) means it's decreasing, and a zero \(d\) means all terms are the same.
• To find \(d\) when given two terms of the sequence, you would subtract the earlier term from the later term and divide by the number of terms between them.

Applying these principles, in our problem, the common difference was found by knowing the 5th and 12th terms and solving the equation: \(7d = 28\), which results in \(d = 4\).

n-th Term Formula of an Arithmetic Sequence

The n-th term formula is a powerful tool for finding any term in an arithmetic sequence without listing all the preceding terms. This formula is expressed as: \(a_n = a_1 + (n-1) \cdot d\).
This formula uses:

• \(a_n\) which is the n-th term you want to find.
• \(a_1\) which is the first term of the sequence.
• \(n\) which is the term number or position in the sequence.
• \(d\) which is the common difference between the terms.

By substituting known values into this formula, you can easily find unknown terms.
In the original problem, we used the n-th term formula twice to set up equations for the 5th and 12th terms. This gave us vital information to find \(a_1\) and \(d\).

Solving Equations to Find Terms

Equation solving is a critical skill when working with arithmetic sequences, especially when some values are missing. Once you have set up equations using the n-th term formula, you can solve them to find the unknowns.
In our original problem, we wanted to find the first term \(a_1\). We began with equations based on known terms:

• For the 5th term: \(a_1 + 4d = 14\)
• For the 12th term: \(a_1 + 11d = 42\)

To isolate \(d\), we subtracted the first equation from the second. This clever step eliminated \(a_1\) and left us an equation with only \(d\), allowing us to easily solve for \(d = 4\). With \(d\) known, it was straightforward to substitute back into one of the original equations and solve for \(a_1 = -2\).
This process highlights the power of algebraic manipulation in solving sequence problems.