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Determine whether the sequence is arithmetic or not. If it is, find the common difference. $$\frac{1}{3}, \frac{2}{3}, \frac{3}{3}, \frac{4}{3}, \frac{5}{3}, \ldots$$

Short Answer

Expert verified
Answer: Yes, the given sequence is arithmetic, and the common difference is \(\frac{1}{3}\).

Step by step solution

01

Identify the given sequence

The given sequence is: $$\frac{1}{3}, \frac{2}{3}, \frac{3}{3}, \frac{4}{3}, \frac{5}{3}, \ldots$$
02

Find the difference between consecutive terms

Let's find the difference between the consecutive terms of the sequence, starting with the first two terms: $$\frac{2}{3} - \frac{1}{3} = \frac{2-1}{3} = \frac{1}{3}$$ Now let's find the difference between the next pair of terms: $$\frac{3}{3} - \frac{2}{3} = \frac{3-2}{3} = \frac{1}{3}$$ We can continue this process for the other terms: $$\frac{4}{3} - \frac{3}{3} = \frac{4-3}{3} = \frac{1}{3}$$ $$\frac{5}{3} - \frac{4}{3} = \frac{5-4}{3} = \frac{1}{3}$$
03

Check if the differences are constant

From Step 2, we can see that the differences we found between consecutive terms are as follows: $$\frac{1}{3}, \frac{1}{3}, \frac{1}{3}, \frac{1}{3}, \ldots$$ Since the differences between the terms are the same, we can conclude that the given sequence is an arithmetic sequence.
04

Identify the common difference

The common difference in an arithmetic sequence is the constant difference between consecutive terms. In this case, the common difference is: $$\frac{1}{3}$$ In conclusion, the given sequence is arithmetic, and its common difference is \(\frac{1}{3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Common Difference
In arithmetic sequences, the **common difference** is a vital concept. It is the constant value that you add to each term in the sequence to obtain the next term. To find this difference, simply subtract any term from the term that follows it. For example, in the sequence \( \frac{1}{3}, \frac{2}{3}, \frac{3}{3}, \ldots \), the common difference can be calculated as follows:
  • From \( \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \)
  • From \( \frac{3}{3} - \frac{2}{3} = \frac{1}{3} \)
What's important to note is this difference remains constant for all consecutive terms, confirming the sequence's arithmetic nature. Calculate the difference consistently, and if it remains the same across all pairs of terms, you've established the common difference.
Consecutive Terms
Understanding **consecutive terms** in a sequence helps in identifying the common difference and understanding the pattern in the sequence. Consecutive terms are simply terms that follow one another without gaps in an ordered list.

Consider the sequence \( \frac{1}{3}, \frac{2}{3}, \frac{3}{3}, \ldots \). Each term directly follows its predecessor in this sequence, with no interruptions or missing numbers. The importance of consecutive terms is that they allow you to easily compute the common difference by a simple subtraction operation.
  • Find the next term by adding the common difference to the current term.
  • These small steps make the analysis much more manageable, showing a clear pattern across the sequence.
Consecutive terms form the backbone of sequence analysis in precalculus and beyond, providing an easy way to track the progression of terms.
Precalculus
**Precalculus** serves as a bridge connecting fundamental mathematical concepts with more complex calculus ideas. It involves the study of different types of sequences, including arithmetic sequences, which prepares students for calculus concepts.

This branch of mathematics equips learners with essential skills needed for higher-level problem-solving. Arithmetic sequences, for instance, involve finding a common difference, understanding term progression, and applying formulas for sequence analysis. Precalculus uses these sequences to model real-world situations, allowing students to grasp mathematical patterns and their implications.
  • Engaging with sequences builds a strong basis for understanding limits and series in calculus.
  • Concentration on sequences sharpens numeracy skills and logical reasoning.
By mastering arithmetic sequences in precalculus, students lay a strong foundation for future mathematical success.
Sequence Analysis
**Sequence Analysis** is the systematic examination of terms in a sequence to uncover essential characteristics. It involves determining whether a sequence is arithmetic, identifying the common difference, and performing further evaluations if necessary.

In the case of arithmetic sequences, sequence analysis helps determine the regularity and pattern, facilitating predictions about future terms.
  • Confirming the same common difference helps establish the sequence is arithmetic.
  • Once this analysis is performed, it becomes easier to generate any term in the sequence by using the formula for the nth term: \( a_n = a_1 + (n-1) \times d \), where \( a_n \) is the nth term, \( a_1 \) is the first term, and \( d \) is the common difference.
Comprehensive sequence analysis aids not just in identifying the type of sequence, but also in understanding the overall structure and behavior of the pattern, useful for both mathematical and applied contexts.

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Most popular questions from this chapter

In Exercises \(49-54,\) you are asked to find a geometric sequence. In each case, round the common ratio r to four decimal places. The amount spent per person per year on cable and satellite TV can be approximated by a geometric sequence \(\left\\{b_{n}\right\\}\) where \(n=1\) corresponds to \(2001 .^{\\#}\) (a) If \(\$ 204.74\) was spent in 2001 and \(\$ 232.22\) was spent in \(2003,\) find a formula for \(b_{n}\) (b) Find the total that will be spent per person from 2001 to 2009 (inclusive).

Find the third and the sixth partial sums of the sequence. $$\left\\{2^{n}\left(2-n^{2}\right)\right\\}_{n \geq 0}$$

In Exercises \(49-54,\) you are asked to find a geometric sequence. In each case, round the common ratio r to four decimal places. Data from the U.S. Centers for Disease Control and Prevention indicate that the number of newly reported cases of AIDs each year can be approximated by a geometric sequence \(\left\\{a_{n}\right\\},\) where \(n=1\) corresponds to 2000 (a) If there were 40,758 cases reported in 2000 and 41,573 cases reported in \(2001,\) find a formula for \(a_{n}\) (b) About how many cases were reported in \(2004 ?\) (c) Find the total number of cases reported from 2000 to 2007 (inclusive).

In Exercises \(43-48,\) find the sum. $$\sum_{n=1}^{5} 5 \cdot 3^{n-1}$$

Deal with the Fibonacci sequence \(\left\\{a_{n}\right\\}\) that was discussed in Example 6. Show that \(\sum_{n=1}^{k} a_{n}=a_{k+2}-1 .\left[\text {Hint: } a_{1}=a_{3}-a_{2}\right.\) \(\left.a_{2}=a_{4}-a_{3} ; \text { etc. }\right]\).

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