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Chapter 6: Problem 34

Find three geometric means between 3 and 9\(\frac{13}{27}\) .

Short Answer

Expert verified
The three geometric means are found using the common ratio \(r = \sqrt[4]{\frac{244}{81}}\): calculate \(G_1 = 3r\), \(G_2 = 3r^2\), \(G_3 = 3r^3\).

Step by step solution

01

Understand the Problem

To find three geometric means between two numbers, we are essentially looking for four numbers forming a geometric sequence: the first number, the three means, and the last number. A geometric sequence is a set of numbers where each term after the first is found by multiplying the previous term by a constant called the common ratio.
02

Setup the Geometric Sequence

Let's denote the terms of the sequence as: 3, \(G_1\), \(G_2\), \(G_3\), and \(9\frac{13}{27}\) (the given fraction can also be written as \(\frac{244}{27}\)). We define the common ratio as \(r\), so the sequence can be expressed as: \(3, 3r, 3r^2, 3r^3, 3r^4 = 9\frac{13}{27}\).
03

Formulate the Equation for Common Ratio

Given the sequence, the equation we need to solve is: \(3r^4 = \frac{244}{27}\). Divide through by 3 to isolate \(r^4\): \(r^4 = \frac{244}{81}\).
04

Solve for the Common Ratio

Now we need to solve for \(r\) by taking the fourth root of both sides of the equation. Hence, \(r = \sqrt[4]{\frac{244}{81}}\). Use a calculator to find \(r\).
05

Calculate the Geometric Means

With the common ratio \(r\) determined, calculate the geometric means: \(G_1 = 3r\), \(G_2 = 3r^2\), \(G_3 = 3r^3\). These are the desired values of the three geometric means.

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Key Concepts

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

Common Ratio
In a geometric sequence, the term "common ratio" refers to the constant factor that you multiply each term by to get the next term. This ratio plays a crucial role in defining the sequence. For instance, consider a simple example: if you have a sequence starting with 2, 6, 18, and so on, the common ratio here is 3 because you multiply each term by 3 to get the next term.
  • To find the common ratio, you can divide any term in the sequence by its preceding term.
  • Mathematically, the formula to calculate the common ratio \( r \) from terms \( a_n \) and \( a_{n-1} \) is: \[ r = \frac{a_n}{a_{n-1}} \]
In the given exercise, calculating the common ratio was essential to locate the three geometric means between 3 and \( 9\frac{13}{27} \). The problem asks to solve for \( r \) using the fourth root, illustrating its fundamental importance.
Geometric Means
Geometric means fill the values in a geometric sequence between two numbers, maintaining the common ratio. When you say you're looking for geometric means, it means you need to find intermediate values in such a sequence. Suppose you're bridging two known numbers in a geometric sequence like 3 and 81.
  • For example, if you find one geometric mean between 3 and 27, it makes the sequence: 3, 9, 27. Here, 9 is the geometric mean because it continues the progression.
  • To find multiple geometric means, assume terms between the starting and ending values, and apply the common ratio to solve for them.
In the exercise, three geometric means were calculated using the common ratio derived to fit between 3 and \( 9\frac{13}{27} \). These intermediate numbers maintain the sequence's integrity, ensuring each follows from the other when multiplied by the common ratio.
Fourth Root
Taking a fourth root is a mathematical operation involving finding a number which, when raised to the power of four, equals the original number. In simpler terms, it reverses the process of raising to the power of four.
  • For example, if you have \( x^4 = 81 \), then \( x = \sqrt[4]{81} = 3 \), because \( 3^4 = 81 \).
  • Calculating a fourth root is usually straightforward with a calculator, crucial in determining the common ratio here.
In the given solution, we needed to find the fourth root of \( \frac{244}{81} \) to calculate the common ratio \( r \). By solving \( r = \sqrt[4]{\frac{244}{81}} \), this illustrates the application of fourth roots in finding values in geometric sequences. This key step ensured that each term maintained the sequence's proportional integrity.

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Most popular questions from this chapter

What is the 9 th term of the geometric sequence \(125,25,5, \ldots ?\)

Use the graphing calculator to evaluate the following series to the nearest hundredth: $$ \begin{array}{llll}{\text { (1) } \sum_{n=1}^{50}\left(1+\frac{1}{n}\right)} & {\text { (2) } \sum_{k=1}^{18} \frac{5}{1+k}} & {\text { (3) } \sum_{n=1}^{20} \frac{(n-1)(-1)^{n}}{n}}\end{array} $$

In \(3-14,\) find the sum of \(n\) terms of each geometric series. $$ \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+a_{n}, n=6 $$

In \(31-39,\) write the first five terms of each sequence. $$ a_{1}=1, a_{n}=2 a_{n-1}+1 $$

An infinitely repeating decimal is an infinite geometric series. Find the rational number represented by each of the following infinitely repeating decimals. \(0.121212 \ldots=0.12+0.0012+0.000012+\cdots\)
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