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Chapter 2: Problem 31

For the following exercises, use the formula given to solve for the required value. \(S u m=\frac{1}{1-r}\) is the formula for an infinite series sum. If the sum is \(5,\) find \(r\).

Short Answer

Expert verified
\( r = \frac{4}{5} \).

Step by step solution

01

Understand the Formula

We are given the formula for the sum of an infinite geometric series: \( S = \frac{1}{1 - r} \). This formula will be used to find the value of \( r \) when the sum is provided.
02

Substitute the Given Value

The problem states that the sum \( S \) of the series is 5. Substitute \( S = 5 \) into the formula: \( 5 = \frac{1}{1 - r} \).
03

Isolate the Denominator

To isolate the denominator \( 1 - r \), take the reciprocal of both sides: \( 1 - r = \frac{1}{5} \).
04

Solve for r

Rearrange \( 1 - r = \frac{1}{5} \) to solve for \( r \). Subtract \( \frac{1}{5} \) from 1 to find \( r \): \( r = 1 - \frac{1}{5} \).
05

Simplify the Expression

Calculate \( 1 - \frac{1}{5} \): Convert 1 to \( \frac{5}{5} \) and subtract \( \frac{1}{5} \): \( \frac{5}{5} - \frac{1}{5} = \frac{4}{5} \). Thus, \( r = \frac{4}{5} \).

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Key Concepts

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

Geometric Series
A geometric series is a sum of terms in a sequence where each term is a constant multiple of the previous term. This constant multiple is known as the common ratio, denoted as \( r \). In a geometric series, the first term is often called \( a \), and the series takes the form \( a, ar, ar^2, ar^3, \ldots \). For example, in the series \( 2, 4, 8, 16, \ldots \), the common ratio \( r \) is 2 because each term is obtained by multiplying the previous term by 2.
Below are some important characteristics of geometric series:
  • Each term increases (or decreases) by the same multiplicative factor \( r \).
  • The series is a type of exponential growth or decay, depending on whether \( r \) is greater or less than 1.
  • Geometric series can be finite or infinite, depending on whether there are limited or unlimited terms.
Understanding geometric series is essential for grasping more advanced topics like infinite geometric series and finding sums.
Infinite Geometric Series
An infinite geometric series is a series where the terms continue indefinitely. When the common ratio \( r \) has an absolute value less than 1 (\(|r| < 1\)), the series has a finite sum, meaning it converges. The sum of an infinite geometric series can be calculated using the formula \( S = \frac{a}{1 - r} \), where \( a \) is the first term and \( r \) is the common ratio.
Key features of infinite geometric series include:
  • Convergence occurs when \(|r| < 1\). In this case, the terms become smaller and smaller, getting close to zero, allowing the series to have a finite sum.
  • Divergence happens when \(|r| \geq 1\). Here, the terms do not approach zero, so the series' sum grows indefinitely.
  • The formula \( S = \frac{a}{1 - r} \) applies only when the series converges, making it necessary to first verify \(|r| < 1\) before using it.
Understanding this concept is crucial when solving problems that involve infinite sums, such as determining the sum given a particular common ratio.
Solving for r
When given the sum of an infinite geometric series and asked to find \( r \), we need to manipulate the formula \( S = \frac{1}{1 - r} \). This formula becomes \( S = \frac{a}{1 - r} \) if the first term \( a \) is not 1.
The steps to solve for \( r \) include:
  • Identify the given sum \( S \). If \( a = 1 \), then use \( S = \frac{1}{1 - r} \).
  • Substitute the known sum into the equation.
  • Rearrange the equation to solve for \( r \). This might involve taking reciprocals or rearranging terms.
  • Simplify to find the exact value of \( r \).
For instance, if the sum of an infinite series is 5 and \( a = 1 \), plug \( S = 5 \) into \( 5 = \frac{1}{1 - r} \), solve \( 1 - r = \frac{1}{5} \), and then solve \( r = 1 - \frac{1}{5} = \frac{4}{5} \). Solving for \( r \) helps to identify how each term scales from the previous one.

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