Problem 17 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 13-18: Find the specified term of the arithmetic sequence that has the two given terms. $$ a_{15} ; \quad a_{3}=7, \quad a_{20}=43 $$

Step-by-Step Solution

Answer

The 15th term \( a_{15} \) is approximately 32.

Step 1: Understanding the Problem

We need to find the 15th term \( a_{15} \) of the arithmetic sequence given \( a_3 = 7 \) and \( a_{20} = 43 \). An arithmetic sequence has a common difference \( d \) between consecutive terms.

Step 2: Derive the General Formula

In an arithmetic sequence, the \( n \)-th term is given by \( a_n = a_1 + (n-1)d \), where \( a_1 \) is the first term, and \( d \) is the common difference. Using \( a_3 \), the equation becomes: \( a_3 = a_1 + 2d = 7 \).

Step 3: Set Up Another Equation for a_20

Similarly, for the 20th term, the formula is \( a_{20} = a_1 + 19d = 43 \). Now we have two equations: \( a_1 + 2d = 7 \) and \( a_1 + 19d = 43 \).

Step 4: Solve the System of Equations

Subtract the first equation from the second: \((a_1 + 19d) - (a_1 + 2d) = 43 - 7\). Simplify to find \( 17d = 36 \). Hence, \( d = \frac{36}{17} \).

Step 5: Find a_1

Substitute \( d = \frac{36}{17} \) into the equation \( a_1 + 2d = 7 \): \( a_1 + 2\left(\frac{36}{17}\right) = 7 \). Solve for \( a_1 \): \( a_1 = 7 - \frac{72}{17} = \frac{47}{17} \).

Step 6: Calculate the 15th Term

Use the general formula to find \( a_{15} = a_1 + 14d \). Substitute \( a_1 = \frac{47}{17} \) and \( d = \frac{36}{17} \): \( a_{15} = \frac{47}{17} + 14\left(\frac{36}{17}\right) = \frac{47}{17} + \frac{504}{17} = \frac{551}{17} = 32.4118 \approx 32 \).

Key Concepts

Common Difference in Sequences

One of the most important aspects of an arithmetic sequence is its **common difference**. This is a constant value that separates consecutive terms in the sequence. The common difference is denoted by the letter \( d \). It can be calculated by subtracting one term from the term that follows it. For example:

• If you have a sequence: 2, 5, 8, 11..., the common difference \( d \) is 5 - 2 = 3.
• Likewise, for a sequence like 10, 7, 4, 1..., the common difference is \( 7 - 10 = -3 \).

To better grasp this, always remember:

• A positive \( d \) means the sequence is increasing.
• A negative \( d \) indicates a decreasing sequence.
• A zero \( d \) means all terms are the same.

By understanding the common difference, you can predict future terms of the sequence easily. In our exercise, we found \( d \) by using given terms \( a_3 \) and \( a_{20} \), calculated as \( \frac{36}{17} \). This helps to establish the full arithmetic sequence.

Terms of a Sequence

Each position in a sequence is called a **term**, and each term can be identified by its position in the sequence, like the 1st term \( a_1 \), 2nd term \( a_2 \), and so on. In the equation for an arithmetic sequence, \( a_n = a_1 + (n-1)d \), each part has a purpose:

• \( a_n \) is the \( n \)-th term we want to find.
• \( a_1 \) is the first term of the sequence.
• \( (n-1) \) shows how many times we add the common difference \( d \) to reach \( a_n \).

This formula is like a map showing how to get from the first term to any term in the sequence. By substituting known values, you can discover any term. For instance, in our task, finding \( a_{15} \) involves substituting the determined values of \( a_1 \) and \( d \) into the formula, resulting in \( a_{15} = 32.412 \).

Solving Systems of Equations

To unravel the secrets of the sequence, sometimes you'll need to use systems of equations. This means you have two or more equations that need to be solved together. In our problem, we had:

• \( a_1 + 2d = 7 \)
• \( a_1 + 19d = 43 \)

These two equations are key to finding both the first term \( a_1 \) and the common difference \( d \). The trick is often to eliminate one of the variables by combining the equations. Here, - Subtracting the first equation from the second eliminated \( a_1 \), simplifying our work.This can make what seems like a complicated problem more manageable. Once \( d \) was found to be \( \frac{36}{17} \), substituting back allowed us to calculate \( a_1 \). Thus, solving these equations unveiled crucial details about the sequence.