Problem 34 - Precalculus: Mathematical for Calculus

Question

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

Step-by-Step Solution

Answer

The 20th term of the sequence is 48.

Step 1: Understand the Arithmetic Sequence Formula

The formula for the n-th term of an arithmetic sequence is given by \( a_n = a_1 + (n-1) \cdot d \), where \( a_n \) is the n-th term, \( a_1 \) is the first term, and \( d \) is the common difference.

Step 2: Set Up Equations Using Given Terms

We know the 12th term, \( a_{12} = 32 \), and the 5th term, \( a_5 = 18 \). These give us two equations: \( a_1 + 11d = 32 \) and \( a_1 + 4d = 18 \).

Step 3: Solve for the Common Difference (d)

Subtract the second equation from the first: \((a_1 + 11d) - (a_1 + 4d) = 32 - 18\). This simplifies to \(7d = 14\), so the common difference \(d = 2\).

Step 4: Solve for the First Term (a_1)

Substitute \(d = 2\) back into one of the equations, e.g., \(a_1 + 4d = 18\). Plugging in \(d\), we get \(a_1 + 8 = 18\). Thus, \(a_1 = 10\).

Step 5: Find the 20th Term

Using the formula \( a_n = a_1 + (n-1) \cdot d \), find \(a_{20}\). We have \(a_1 = 10\), \(d = 2\), and \(n = 20\). So, \(a_{20} = 10 + (20-1) \cdot 2 = 10 + 38 = 48\).

Key Concepts

Common Difference

In an arithmetic sequence, the **common difference** is a crucial element. It's the constant amount we add or subtract between consecutive terms in the sequence. This difference is what makes an arithmetic sequence quite straightforward but also very systematic. It's key to determining each term's value with consistency.
For example, if our sequence starts with a number and the common difference is 2, each subsequent number is simply 2 more than the last. If we have numbers like 3, 5, 7, 9, and so on, we can see the common difference here is 2.

• Understanding it helps in constructing the entire sequence from just one known term and the difference.
• It plays a central role in building the sequence equations used to solve problems.
• Once identified, the common difference helps us find any term we want by plugging it into formulae.

Being especially useful, the common difference allows us also to verify if a sequence really is arithmetic. If you compute the difference between consecutive terms and it's identical each time, bingo—an arithmetic sequence!

n-th Term Formula

The **n-th term formula** is the equation we use to find any specific term in our arithmetic sequence. It's like having a magic key that opens the door to any position within the sequence. The formula is:\[ a_n = a_1 + (n-1) \cdot d \]
Here:

• \(a_n\) is the term you're aiming to find.
• \(a_1\) is the very first term in the sequence.
• \(d\) is our trusty common difference.
• \(n\) stands for the term's position in the sequence.

Knowing even just the first term and the common difference, you can leap around your sequence, calculating any term you fancy.
This formula unifies the entire structure, allowing prediction or back-calculation into any sequence position. Quite powerful, especially when dealing with large term numbers, like wanting to know, say, the 100th term.

Sequence Equations

**Sequence equations** are the means by which we articulate relationships between terms in arithmetic sequences. Using known terms and seeking unknowns, we can establish equations that pin down our arithmetic sequence perfectly.
In the exercise you've seen, two sequence equations were formed:
\[a_1 + 11d = 32\]
\[a_1 + 4d = 18\]
These are simply versions of our n-th term formula, set up with given term values and the reasoning process to solve them.

• Such equations help us isolate variables and unknowns.
• They are great for solving sequences where a couple of terms are known, allowing us to uncover hidden elements like the first term and common difference.
• We use them in sequence exercises to guide a systematic approach to find missing terms.

These equations are invaluable tools that get us from a few bits of information to a complete understanding of the sequence structure. Solving them usually involves recognizing patterns in the numbers and relying on algebraic skills.