Problem 23 - A Graphical Approach to Precalculus with Limits

Question

Find \(a_{8}\) and \(a_{n}\) for each arithmetic sequence. $$a_{3}=\pi+2 \sqrt{e}, a_{4}=\pi+3 \sqrt{e}$$

Step-by-Step Solution

Answer

\(a_8 = \pi + 7\sqrt{e}\), \(a_n = \pi + (n-1)\sqrt{e}\).

Step 1: Understand the Problem

We are given two terms of an arithmetic sequence: \( a_3 = \pi + 2\sqrt{e} \) and \( a_4 = \pi + 3\sqrt{e} \). We need to find \( a_8 \) and the general term \( a_n \) for the sequence.

Step 2: Find the Common Difference

In an arithmetic sequence, the difference between consecutive terms is constant. Thus, the common difference \( d \) can be found as follows:\[ d = a_4 - a_3 = (\pi + 3\sqrt{e}) - (\pi + 2\sqrt{e}) = \sqrt{e}. \]

Step 3: Determine the Initial Term

To find the first term \( a_1 \), we use the formula for the \( n \)-th term in an arithmetic sequence, \( a_n = a_1 + (n-1)d \). Using the known term \( a_3 \), we have:\[ a_3 = a_1 + 2d = \pi + 2\sqrt{e}. \]Substituting \( d = \sqrt{e} \) yields:\[ a_1 + 2\sqrt{e} = \pi + 2\sqrt{e}, \]which simplifies to \( a_1 = \pi. \)

Step 4: Write the Formula for the General Term

With \( a_1 = \pi \) and \( d = \sqrt{e} \), the general formula for the \( n \)-th term is:\[ a_n = a_1 + (n-1)d = \pi + (n-1)\sqrt{e}. \]

Step 5: Find the 8th Term

Finally, use the general formula to find \( a_8 \):\[ a_8 = \pi + (8-1)\sqrt{e} = \pi + 7\sqrt{e}. \]

Key Concepts

Common Difference

In any arithmetic sequence, one fundamental property is the **common difference**. This is the consistent interval between any two successive terms. Imagine stepping up a staircase where each step is the same height. Similarly, the common difference represents this step, ensuring each term in the sequence increments regularly.

To find this difference from given terms, you simply subtract an earlier term from a later term. Here, we have the third term as \( a_3 = \pi + 2\sqrt{e} \) and the fourth term as \( a_4 = \pi + 3\sqrt{e} \). By subtracting \( a_3 \) from \( a_4 \), we calculate:

• \( d = a_4 - a_3 \)
• \( d = (\pi + 3\sqrt{e}) - (\pi + 2\sqrt{e}) \)
• \( d = \sqrt{e} \)

This simple yet pivotal calculation shows the linear increase from one term to the next.

General Term Formula

The **General Term Formula** is essential for understanding the structure and pattern of an arithmetic sequence. It allows you to determine any term's value based on its position number.

The formula is expressed as \( a_n = a_1 + (n-1)d \), where

• \( a_n \) represents the \( n \)-th term,
• \( a_1 \) is the first term of the sequence,
• \( d \) is the common difference, and
• \( n \) is the specific position of the term you are looking for.

By substituting the known values, you not only find specific terms, but also gain insights into the behavior of the whole sequence.

Arithmetic Sequence Formula

The **Arithmetic Sequence Formula** provides a blueprint for generating the sequence of numbers. This formula uses the first term and the common difference to create a linear pattern of terms. Knowing the first term and how each term relates to its predecessor allows us to construct the entire sequence.

For our specific sequence, we have calculated:

• First term \( a_1 = \pi \)
• Common difference \( d = \sqrt{e} \)

Therefore, the arithmetic sequence formula becomes:

• \( a_n = \pi + (n-1)\sqrt{e} \)

This equational framework helps predict and calculate numerous values across the sequence, ensuring you always have the tools to solve for unknowns.

Term Calculation

**Term Calculation** involves using the general term formula to find the value of a specific term in an arithmetic sequence. This process connects the dots between knowing the formula and applying it practically.

For illustration, let's calculate \( a_8 \), the eighth term of our sequence. Using our derived formula \( a_n = \pi + (n-1)\sqrt{e} \), substitute \( n = 8 \):

• \( a_8 = \pi + (8-1)\sqrt{e} \)
• \( a_8 = \pi + 7\sqrt{e} \)

Such calculations allow you to determine the value of any term directly, without having to sequentially calculate preceding terms. This method illustrates the efficiency and versatility of understanding arithmetic sequences.