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Chapter 10: Problem 37

For each arithmetic sequence described, find \(a_{1}\) and \(d\) and construct the sequence by stating the general, or \(n\)th, term. The 4th term is 3 and the 22nd term is \(15 .\)

Short Answer

Expert verified
First term \( a_1 = 1 \), common difference \( d = \frac{2}{3} \), sequence's nth term is \( a_n = 1 + (n-1) \cdot \frac{2}{3} \).

Step by step solution

01

Identify given terms

We are given that the 4th term, \( a_4 = 3 \), and the 22nd term, \( a_{22} = 15 \). The goal is to find the first term \( a_1 \) and the common difference \( d \).
02

General term formula

The general term for an arithmetic sequence is given by \( a_n = a_1 + (n-1) \cdot d \). We will use this formula to express the given terms in terms of \( a_1 \) and \( d \).
03

Set up equations from given terms

Using the general term formula:1. For the 4th term, \( a_4 = a_1 + 3d = 3 \).2. For the 22nd term, \( a_{22} = a_1 + 21d = 15 \).
04

Solve for common difference \(d\)

Subtract the first equation from the second:\[(a_1 + 21d) - (a_1 + 3d) = 15 - 3\]This simplifies to:\[18d = 12\]Solve for \(d\):\[d = \frac{12}{18} = \frac{2}{3}\]
05

Solve for first term \(a_1\)

Substitute \( d = \frac{2}{3} \) back into the equation for the 4th term:\[a_1 + 3 \cdot \frac{2}{3} = 3\]This simplifies to:\[a_1 + 2 = 3\]Solve for \(a_1\):\[a_1 = 3 - 2 = 1\]
06

Construct the sequence formula

Now that we have both \( a_1 = 1 \) and \( d = \frac{2}{3} \), we can construct the general term of the sequence:\[a_n = 1 + (n-1) \cdot \frac{2}{3}\]Simplify it if necessary for the nth term expression.

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Key Concepts

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

Common Difference
In an arithmetic sequence, the common difference is a fixed number that is added to each term to get to the next term. This concept is crucial because it defines the pattern of the sequence. For example, in our exercise, we had to determine the common difference, which turned out to be \(\frac{2}{3}\).

To find the common difference:
  • Identify two known terms in the sequence and their positions, like the 4th term \(a_4 = 3\), and the 22nd term \(a_{22} = 15\).
  • Use the general term formula for each: \(a_n = a_1 + (n-1) \cdot d\). Substitute and set up equations with these terms.
  • Subtract one equation from the other to eliminate \(a_1\) and solve for \(d\).


Handling the arithmetic correctly is important to accurately determine \(d\), especially when it involves fractions like it did here with \(\frac{2}{3}\). Understanding the common difference helps in predicting any term in the sequence precisely.
General Term Formula
The general term formula of an arithmetic sequence is a mathematical expression used to determine any term in the sequence. It is given by \(a_n = a_1 + (n-1) \cdot d\).
This formula helps us understand how an arithmetic sequence behaves and allows us to find any term without constructing the entire sequence.

Using the formula:
  • \(a_1\) is the first term of the sequence.
  • \(d\) is the common difference between consecutive terms.
  • \(n\) is the position of the term you want to find.


By manipulating this formula according to the information given about particular terms, we can set up equations to find unknown values. In the exercise, setting equations for the 4th and 22nd terms allowed us to find both \(a_1\) and \(d\).

This formula is powerful for quickly coming up with any desired term which is particularly useful if we are looking into very large sequences.
Arithmetic Progression
An arithmetic progression, also known as an arithmetic sequence, is a list of numbers where each term after the first is found by adding a constant, called the common difference, to the previous term.

Key Characteristics:
  • It is defined just by two elements: the first term and the common difference.
  • The sequence grows linearly, as each term increases or decreases by the same amount.


In the given exercise, we constructed an arithmetic progression where the sequence starts at \(1\) (this is our \(a_1\)), and we add \(\frac{2}{3}\) to get each subsequent term.

This means the sequence is: \[1, \frac{5}{3}, \frac{7}{3}, 3, \ldots\] Understanding arithmetic progression goes a long way in solving various kinds of mathematical problems, not just in terms of direct calculation but also in understanding growth, patterns, and behavior of numbers over iterations.

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Most popular questions from this chapter

Several of the results studied in calculus must be proved by mathematical induction. Apply mathematical induction to prove each formula. $$\begin{array}{l}(\pi+1)+(\pi+2)+(\pi+3)+\dots+(\pi+n) \\\=\frac{n(2 \pi+n+1)}{2}\end{array}$$

In calculus, we study the convergence of sequences. A sequence is convergent when its terms approach a limiting value. For example, \(a_{n}=\frac{1}{n}\) is convergent because its terms approach zero. If the terms of a sequence satisfy \(a_{1} \leq a_{2} \leq a_{3} \leq \ldots \leq a_{n} \leq \ldots\) the sequence is monotonic nondecreasing. If \(a_{1} \geq a_{2} \geq a_{3} \geq \ldots \geq a_{n} \geq \ldots,\) the sequence is monotonic nonincreasing. Classify each sequence as monotonic or not monotonic. If the sequence is monotonic, determine whether it is nondecreasing or nonincreasing. $$a_{n}=\frac{2+(-1)^{n}}{n+4}$$

Apply mathematical induction to prove $$\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \cdots\left(1+\frac{1}{n}\right)=n+1$$

In calculus, we study the convergence of geometric series. A gcometric series with ratio \(r\) diverges if \(|r| \geq 1 .\) If \(|r|<1\) then the geometric series converges to the sum \(\sum_{n=0}^{\infty} a r^{n}=\frac{a}{1-r}\) Determine the convergence or divergence of the series. If the series is convergent, find its sum. $$\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}+\dots$$

The Fibonacci sequence is defined by \(a_{1}=1, a_{2}=1,\) and \(a_{n}=a_{n-2}+a_{n-1}\) for \(n \geq 3 .\) The ratio \(\frac{a_{n+1}}{a_{n}}\) is an approximation of the golden ratio. The ratio approaches a constant \(\phi\) (phi) as \(n\) gets large. Find the golden ratio using a graphing utility.
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