Problem 34 - Algebra and Trigonometry

Question

The 12th term of an arithmetic sequence is \(32,\) and the fifth term is \(18 .\) Find the 20th term.

Step-by-Step Solution

Answer

The 20th term is 48.

Step 1: Identify the Sequence Formula

For an arithmetic sequence, the nth term can be found using the formula: \( a_n = a_1 + (n-1) \cdot d \), where \( a_1 \) is the first term and \( d \) is the common difference.

Step 2: Setup Equations for Given Terms

Use the formula to express the given terms in equations. \( a_{12} = 32 \) gives us: \( a_1 + 11d = 32 \). Similarly, \( a_5 = 18 \) gives us: \( a_1 + 4d = 18 \).

Step 3: Solve the System of Equations

Subtract the equation \( a_1 + 4d = 18 \) from \( a_1 + 11d = 32 \) to eliminate \( a_1 \): \[ (a_1 + 11d) - (a_1 + 4d) = 32 - 18 \]. Simplifying gives \( 7d = 14 \), so \( d = 2 \).

Step 4: Find the First Term

Substitute \( d = 2 \) back into one of the equations, say \( a_1 + 4d = 18 \): \[ a_1 + 4 \cdot 2 = 18 \]. Simplifying gives \( a_1 = 10 \).

Step 5: Calculate the 20th Term

Use the formula for the nth term to find \( a_{20} \): \[ a_{20} = a_1 + 19d \]. Substitute \( a_1 = 10 \) and \( d = 2 \): \[ a_{20} = 10 + 19 \cdot 2 = 10 + 38 = 48 \].

Key Concepts

Understanding the nth Term Formula

The nth term formula is a valuable tool used to find any term in an arithmetic sequence. An arithmetic sequence is a list of numbers with a constant difference between consecutive terms. The formula is:

• \( a_n = a_1 + (n-1)\cdot d \)

where \( a_n \) is the nth term, \( a_1 \) is the first term, and \( d \) is the common difference.
By knowing the position, \( n \), within the sequence and the first term, we can easily calculate the nth term if we also know the common difference.
This formula simplifies the process of sequence problem-solving, allowing us to explore different positions in the sequence without generating all previous terms.

Identifying the Common Difference

The common difference, denoted as \( d \), is crucial in defining an arithmetic sequence. It is the fixed amount added to each term to find the next term.

• Difference is found by subtracting any two consecutive terms.
• For consistent sequences, this value remains the same throughout.

For example, in the given problem, the difference between the 12th term (\(32\)) and the fifth term (\(18\)) can be determined once you set your equations using the nth term formula.
Using equations such as \( a_1 + 11d = 32 \) and \( a_1 + 4d = 18 \), subtracting them provides \( 7d = 14 \), thus \( d = 2 \). This consistent difference allows us to predict and calculate any future or previous term in the sequence effectively.

Solving with a System of Equations

Systems of equations are often used when dealing with problems involving sequences. Here, we set up simultaneous equations using the given terms in the sequence and the nth term formula.

• Equation 1: \( a_1 + 11d = 32 \)
• Equation 2: \( a_1 + 4d = 18 \)

By subtracting one equation from the other, we eliminate one variable. In this instance, eliminating \( a_1 \) gives us:

• \( 7d = 14 \)

Solving for \( d \) provides the common difference. This process exemplifies sequence problem-solving by efficiently incorporating algebraic techniques to find unknowns in arithmetic sequences. It demonstrates how systems of equations can simplify finding components like the common difference or the first term.

Effective Sequence Problem-Solving

Tackling sequence problems systematically aids in achieving precise results. Steps involve:

• Identifying key formulas like the nth term formula.
• Writing equations based on given terms.
• Solving these equations to find unknowns.

Calculating unknown terms requires substituting back into the nth term formula. For instance, the 20th term calculation is critical to this problem-solving approach:

• Substitute \( a_1 = 10 \) and \( d = 2 \) into \( a_{20} = a_1 + 19d \).
• Result: \( a_{20} = 48 \)

Each sequence problem involves understanding the relationship between terms, applying the nth term formula, and efficiently solving any equations derived from known terms. Whether you're predicting future terms or solving for a missing starting value, understanding these relationships is key.