/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 142 Find two different geometric ser... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 142

Find two different geometric series with sums of \( 4 \).

Short Answer

Expert verified
Two geometric series with sums of 4 are 1) 1, 0.5 and 2) 4

Step by step solution

01

First Geometric Series

Let's start by choosing \( a = 1 \), and the common ratio \( r = 0.5 \). We want to find a number of terms such that the sum is \( 4 \). For \( |r|<1 \), we use the formula \( S = \frac{a(1 - r^n)}{1 - r} \). Substituting in our values, we get \( 4 = \frac{1(1 - 0.5^n)}{1 - 0.5} \). Solving this equation for \( n \), we find that \( n = 2 \), so our first series is \( 1, 0.5 \)
02

Second Geometric Series

For the second series, let's choose \( a = 4 \) and \( r = 2 \). This time, because \( |r|>1 \), we use the formula \( S = \frac{a(r^n - 1)}{r - 1} \). Substituting in our values, we get \( 4 = \frac{4(2^n - 1)}{2 - 1} \). Solving this equation for \( n \), we find that \( n = 0 \), so our second series is simply \( 4 \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Understanding the Common Ratio in a Geometric Series
In a geometric series, the common ratio is a key concept. It represents the factor by which each term in the series is multiplied to get the next term.
The common ratio is denoted by the letter \( r \). For example, in a series where each term is half of the previous one, the common ratio is \( 0.5 \). This means if the first term \( a \) is \( 1 \), the second term would be \( 1 \times 0.5 = 0.5 \).
The common ratio can be either greater than \( 1 \) or less than \( 1 \). This affects the behavior of the series.
  • If \(|r| < 1\), the series converges or decreases, meaning the terms get smaller as they progress.
  • If \(|r| > 1\), the series diverges or increases, where terms grow larger.

Understanding this helps determine which formula to use when calculating the sum of a series.
Calculating the Sum of a Geometric Series
Finding the sum of a geometric series depends largely on the value of the common ratio. There are two main scenarios, based on the size of \( r \).
  • When \(|r| < 1\), the sum of the geometric series is calculated using the formula \( S = \frac{a(1 - r^n)}{1 - r} \). This works because the terms get smaller and eventually converge.
  • When \(|r| > 1\), the terms grow, and the sum is calculated with \( S = \frac{a(r^n - 1)}{r - 1} \). Here, terms increase more rapidly.

It's essential to choose the appropriate formula. For example, in a geometric series given \( a = 1 \) and \( r = 0.5 \), we calculate the sum to be \( 4 \) over 2 terms. Meanwhile, if \( a = 4 \) and \( r = 2 \), using the second formula with the sum \( 4 \), the series ends practically after zero terms.
Determining the Number of Terms in a Series
The number of terms in a geometric series, denoted as \( n \), is essential when calculating the sum.
To solve for \( n \), one often rearranges the sum formula for the series. Let's say you want a series whose sum is \( 4 \) and you already have your first term \( a \) and common ratio \( r \).
  • For \(|r| < 1\), solve the equation \( 4 = \frac{a(1 - r^n)}{1 - r} \).
  • For \(|r| > 1\), use \( 4 = \frac{a(r^n - 1)}{r - 1} \).

By substituting known values into these formulas, and solving for \( n \), you determine the number of terms needed. For instance, with \( a = 1 \) and \( r = 0.5 \), you find \( n = 2 \). This method ensures the accurate construction of a series meeting specific sum requirements.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercises 9 - 14, determine the sample space for the experiment. A sales representative makes presentations about a product in three homes per day. In each home, there may be a sale (denote by \( S \)) or there may be no sale(denote by \( F \)).

What is the relationship between \( _nC_r \) and \( _nC_{n - r} \)?

Three people have been nominated for president of a class. From a poll, it is estimated that the first candidate has a \( 37\% \) chance of winning and the second candidate has a \( 44\% \) chance of winning. What is the probability that the third candidate will win?

In Exercises 79 - 86, solve for \( n \). \( _{n + 2} P_3 = 6 \cdot _{n + 2}P_1 \)

In Exercises 39 - 42, you are given the probability that an event will not happen. Find the probability that the event will happen. \( P(E') = 0.92 \)
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.