Problem 27 - Precalculus with Limits

Question

In Exercises 23 - 32, find a formula for for the arithmetic sequence. \( 4, \dfrac{3}{2}, -1, -\dfrac{7}{2}, \cdots \)

Step-by-Step Solution

Answer

The general formula for the given arithmetic sequence is \( a_n = -\dfrac{5n}{2} +\dfrac{13}{2} \).

Step 1: Identify the common difference

Subtract the first term from the second: \( \dfrac{3}{2} - 4 = -\dfrac{5}{2} \). \nSubtract the second term from the third to check if the difference is the same: \( -1 - \dfrac{3}{2} = -\dfrac{5}{2} \). \nThe common difference, \( d \) is \(-\dfrac{5}{2}\)

Step 2: Find the formula for the nth term

Using the formula \( a_n = a_1 + (n-1)*d \), where \( a_1 \) is the first term and \( d \) is the common difference, the formula for these terms of this arithmetic sequence is \( a_n = 4 + (n - 1)*\(-\dfrac{5}{2}\) \)

Step 3: Simplify the expression

Keep 4 as it is and multiply \( (n-1) \) by \( -\dfrac{5}{2} \) which results to \( -\dfrac{5n}{2} +\dfrac{5}{2} \). So, when this is added to 4 we get the formula for the nth term of the sequence \( a_n = -\dfrac{5n}{2} +\dfrac{5}{2} + 4 \), which simplifies to \( a_n = -\dfrac{5n}{2} +\dfrac{13}{2} \).

Key Concepts

Common Difference

In arithmetic sequences, the common difference is a crucial element. It's the consistent interval between consecutive terms of the sequence. To find it, simply subtract one term from the next. In our exercise example, the common difference is calculated by subtracting the first term from the second term:
\( \frac{3}{2} - 4 = -\frac{5}{2} \).
To ensure that the sequence is indeed arithmetic, we verify that this difference is the same for subsequent terms. When we subtract the third term from the second:
\( -1 - \frac{3}{2} = -\frac{5}{2} \),
we find it matches our initial calculation. Thus, the common difference, denoted by \( d \), is \( -\frac{5}{2} \), confirming the properties of an arithmetic sequence.

Nth Term of an Arithmetic Sequence

To predict any term in an arithmetic sequence, we use the formula for the nth term, represented as \( a_n \). This algebraic expression provides a method to calculate the exact value of any term based on its position (n) in the sequence. The formula is given by:
\( a_n = a_1 + (n-1)*d \),
where \( a_1 \) is the first term and \( d \) is the common difference we previously identified. Applying this to our exercise, we substitute with the first term (4) and the common difference (\( -\frac{5}{2} \)):
\( a_n = 4 + (n-1)*\big(-\frac{5}{2}\big) \).
This algebraic expression represents a general rule to find any term in the given sequence.

Sequence and Series

A sequence is a set of numbers listed in a specific order. In mathematics, different types of sequences are defined, and one such type is the arithmetic sequence, as shown in this exercise. Conversely, a series is the sum of the terms of a sequence. Understanding the concept of a sequence is fundamental before one can delve into series since the latter is built upon the former. For arithmetic sequences, the series can be calculated with a different formula, utilizing the first term, the last term, and the total number of terms in the sequence.

It's essential to distinguish between these concepts in order to apply the correct methods for solving problems related to sequences and series.

Algebraic Expressions

Algebraic expressions are integral to formulating mathematical relationships in sequences. In the nth term formula of an arithmetic sequence, the expression consists of constants, variables, and arithmetic operations. Our expression derived in Step 2 of the solution,
\( a_n = 4 + (n-1)*\big(-\frac{5}{2}\big) \),
involves multiplication and addition to relate any term's position \( n \) to its value in the sequence. Simplifying this expression, as shown in Step 3, leads to a more efficient calculation:
\( a_n = -\frac{5}{2}n + \frac{13}{2} \).
This final form is the simplified algebraic expression representing the nth term and highlighting the importance of algebra in understanding and representing arithmetic sequences.