Problem 105 - Precalculus with Limits

Question

In Exercises 105 and 106, determine whether the statement is true or false. Justify your answer. Given an arithmetic sequence for which only the first two terms are known, it is possible to find the \( n \)th term.

Step-by-Step Solution

Answer

The statement is true. Given the first two terms of an arithmetic sequence, it is indeed possible to determine the \( n \)th term using the formula for an arithmetic sequence.

Step 1: Identify the given

The problem states that we are given the first two terms of an arithmetic sequence. Let's denote the first term as \( a \) and the second term as \( b \).

Step 2: Identify the common difference

The common difference is a crucial part of an arithmetic sequence. It can be obtained by simply subtracting the first term from the second term, i.e., \( d = b - a \).

Step 3: Formulate the nth term formula

The general formula for the \( n \)th term of an arithmetic sequence is \( a_n = a + (n-1) * d \). Here, \( n \) is the term number (where \( n \) ≥ 1), \( a \) is the first term and \( d \) is the common difference.

Step 4: Check if n-th term can be calculated

We are supposed to check if the \( n \)th term can be calculated knowing only first two terms and the given sequence is arithmetic. As per the formula in step 3, the \( n \)th term depends on the first term \( a \), the common difference \( d \), and the term number \( n \). We know \( a \) and \( d \) from the first two terms of the sequence. And \( n \) is our choice, so we can find the \( n \)th term.

Key Concepts

Arithmetic Sequence

An arithmetic sequence is a list of numbers with a consistent increment, or 'common difference', between consecutive terms. Think of it like a staircase where each step up or down is evenly spaced from the one before it. This structure creates a predictable pattern, making it simple to identify any term in the sequence knowing just a few starting points. For example, in the sequence 2, 4, 6, 8, ..., each number increases by 2; thus, 2 is the common difference here.

Understanding arithmetic sequences is fundamental because they are not merely a set of numbers but represent a progression that can be seen in real-world phenomena such as time intervals, distances, or financial growth projections. This predictability is what makes arithmetic sequences particularly interesting in the realm of mathematics.

Common Difference

The 'common difference' is the backbone of an arithmetic sequence, determining the unique characteristic of that sequence. It's the steady increase or decrease from one term to the next, and once identified, it unveils the sequence's pattern. To find the common difference, we subtract the first term from the second term. Mathematically, if we denote the first term as 'a' and the second term as 'b', the common difference 'd' is calculated by the simple formula:
\( d = b - a \)

This constant value is crucial for computing subsequent terms of the sequence. It's like having a secret key that unlocks the rest of the sequence, allowing you to leap from one term to any other downstream without having to count each intermediate step.

Sequence Formula

The sequence formula for an arithmetic sequence is a clear-cut way to locate any term within the sequence, regardless of its position. This formula, generally expressed as:
\( a_n = a + (n-1) * d \)
, where \( a_n \) is the nth term we're looking for, 'a' is the first term, 'd' is the common difference, and 'n' is the term's number in the sequence (starting from 1).

The beauty of this formula lies in its sweeping applicability. With it, you're not just inching along from one term to the next; you're making great strides across the sequence, leaping directly to the term you need. This direct route saves time and makes problem-solving much more efficient.

Mathematical Induction

Mathematical induction is an incredibly powerful proof technique that works somewhat like a domino effect. To use it in the context of an arithmetic sequence, you would start by proving that a certain property holds for the first term. You then assume it holds for the nth term and show that because it's true for nth term, it must also be true for the (n+1)th term. This chain of logical steps ensures that, as long as the initial case holds, the property will hold for every following term, no matter how far along the sequence you go.

It's a method of establishing the truth of an infinite number of statements, harnessing the certainty of a base case and the logic of a generic step from n to n+1. In doing so, mathematical induction confirms the building blocks that underpin the entire structure, or sequence, in this case.