Problem 88 - A Graphical Approach to Precalculus with Limits

Question

Find all arithmetic sequences \(a_{1}, a_{2}\) \(a_{3}, \ldots\) such that \(a_{1}^{2}, a_{2}^{2}, a_{3}^{2}, \ldots\) is also an arithmetic sequence.

Step-by-Step Solution

Answer

The sequence must be constant; all terms are equal.

Step 1: Define the Arithmetic Sequence

An arithmetic sequence is defined by its first term \(a_1\) and common difference \(d\). The general term of the sequence is given by \(a_n = a_1 + (n-1)d\).

Step 2: Express Squared Sequence Terms

For the sequence where each term is squared, express the first few terms: \(a_1^2, a_2^2, a_3^2\). Start with: \(a_1^2\), \(a_2^2= (a_1 + d)^2\), and \(a_3^2= (a_1 + 2d)^2\).

Step 3: Setting Up Arithmetic Condition for Squared Sequence

For the sequence \(a_1^2, a_2^2, a_3^2, \ldots\) to be arithmetic, the difference between consecutive terms must be constant. Set up the condition: \(a_2^2 - a_1^2 = a_3^2 - a_2^2\).

Step 4: Expand and Simplify the Condition

Expand the terms: \((a_1 + d)^2 - a_1^2 = (a_1 + 2d)^2 - (a_1 + d)^2\). This becomes \((a_1^2 + 2a_1d + d^2 - a_1^2) = (a_1^2 + 4a_1d + 4d^2 - a_1^2 - 2a_1d - d^2)\).

Step 5: Solve the Equation

Simplify to get \(2a_1d + d^2 = 2a_1d + 3d^2\). Simplifying further gives \(0 = 2d^2\), which implies \(d = 0\).

Step 6: Conclusion for the Arithmetic Sequence

Since \(d = 0\), the original sequence \(a_1, a_2, a_3, \ldots\) must be a constant sequence. Therefore, \(a_1 = a_2 = a_3 = \ldots\).

Step 7: Final Verification

In a constant sequence, each term is identical. Thus, squaring each term results in another constant sequence, which meets the condition of being arithmetic. This confirms that \(a_n = a_1\) for all \(n\).

Key Concepts

Sequences and Series

Arithmetic sequences are a fundamental concept in mathematics, where each term differs from the previous one by a constant value, known as the common difference (denoted as \(d\)). These sequences play a vital role not only in algebra, but also in understanding patterns and predicting future terms in a set. For any arithmetic sequence, given its first term \(a_1\) and a common difference \(d\), the \(n\)-th term is expressed as \(a_n = a_1 + (n-1)d\).
This simple formula allows us to generate the entire sequence once the initial parameters are known. The beauty of arithmetic sequences is their linear quality, which makes them both predictable and manageable.
Series, on the other hand, involve the summation of terms in a sequence. While the focus here is on sequences, understanding the transition to series is valuable, especially when analyzing patterns such as in sequences where terms are squared.

Mathematical Reasoning

Mathematical reasoning is the core skill that allows us to make sense of problems and seek solutions logically. In the problem presented, the reasoning involves understanding that two separate arithmetic sequences need to meet the same criteria: both the original sequence and the sequence of squared terms must be arithmetic. To achieve this, we determine that the difference between consecutive squared terms must remain constant.
Through logical deduction and manipulation, we set up the condition \(a_2^2 - a_1^2 = a_3^2 - a_2^2\).
This shows how mathematical reasoning is about breaking down complex problems into simpler steps by examining relationships and developing conditions systematically. Using reasoning effectively involves testing, expanding, and simplifying assumptions to reach valid conclusions.

Algebraic Manipulation

Algebraic manipulation is the process of transforming and simplifying expressions to make calculations easier or to solve equations. In this exercise, after establishing the arithmetic condition for the squared sequence \((a_2^2 - a_1^2 = a_3^2 - a_2^2)\), the task was to expand and equate it:

• Start by expanding \( (a_1 + d)^2 - a_1^2 \) and \( (a_1 + 2d)^2 - (a_1 + d)^2 \).
• Simplify using algebra – cancel common terms and equate like terms.
• This leads to the simplified equation \(0 = 2d^2\).

The result \(d = 0\) reveals that the common difference is zero, indicating a constant sequence where every term is equal to the first one. Algebraic manipulation remains a crucial technique in solving equations and proof scenarios, ensuring the fluency needed to address increasingly complex problems.