/*! 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 20 Consider the geometric sequence ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 20

Consider the geometric sequence with first term 59,049 and common ratio \(-\frac{1}{3}\). (a) Find the first ten terms and the 33 rd term of this sequence. (b) Find the sum of the first 12 terms.

Short Answer

Expert verified
First 10 terms: 59049, -19683, 6561, -2187, 729, -243, 81, -27, 9, -3. 33rd term: approximately 0. Sum of first 12 terms: about 44314.

Step by step solution

01

Understand the sequence formula

A geometric sequence can be defined using the formula \( a_n = a_1 \, r^{n-1} \), where \( a_n \) is the nth term, \( a_1 \) is the first term, and \( r \) is the common ratio. For this sequence, the first term \( a_1 = 59049 \) and the common ratio \( r = -\frac{1}{3} \).
02

Calculate the first 10 terms

Using the formula \( a_n = a_1 \, r^{n-1} \), calculate each term from 1 to 10:1. \( a_1 = 59049 \, (-\frac{1}{3})^0 = 59049 \)2. \( a_2 = 59049 \, (-\frac{1}{3})^1 = -19683 \)3. \( a_3 = 59049 \, (-\frac{1}{3})^2 = 6561 \)4. \( a_4 = 59049 \, (-\frac{1}{3})^3 = -2187 \)5. \( a_5 = 59049 \, (-\frac{1}{3})^4 = 729 \)6. \( a_6 = 59049 \, (-\frac{1}{3})^5 = -243 \)7. \( a_7 = 59049 \, (-\frac{1}{3})^6 = 81 \)8. \( a_8 = 59049 \, (-\frac{1}{3})^7 = -27 \)9. \( a_9 = 59049 \, (-\frac{1}{3})^8 = 9 \)10. \( a_{10} = 59049 \, (-\frac{1}{3})^9 = -3 \)
03

Find the 33rd term

To find the 33rd term \( a_{33} \), use the formula with \( n=33 \):\( a_{33} = 59049 \, (-\frac{1}{3})^{32} = \frac{59049}{3^{32}} = \frac{59049}{1853020188851841} = 0 \).As the denominator is much larger, the 33rd term effectively rounds to \( 0 \).
04

Find the sum of the first 12 terms

Use the sum formula for the first n terms of a geometric sequence: \( S_n = a_1 \frac{1-r^n}{1-r} \) for \( r eq 1 \). Here, \( n = 12 \), \( a_1 = 59049 \), and \( r = -\frac{1}{3} \):\( S_{12} = 59049 \frac{1-(-\frac{1}{3})^{12}}{1-(-\frac{1}{3})} \).Calculate \( (-\frac{1}{3})^{12} = \frac{1}{531441} \) and substitute:\( S_{12} = 59049 \frac{1-\frac{1}{531441}}{1+\frac{1}{3}} = 59049 \frac{531440}{531441} \, \times \frac{3}{4} \).Thus, \( S_{12} \approx 44314 \).

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.

Common Ratio
In a geometric sequence, the common ratio is the factor by which we multiply each term to get the next term. In our exercise, the common ratio is provided as \( r = -\frac{1}{3} \). This means that to find each subsequent term, you multiply the previous term by \(-\frac{1}{3}\).

The nature of this ratio, being negative and a fraction, means that:
  • Each consecutive term will alternate sign, causing the sequence to switch between positive and negative values.
  • The magnitude of each term will decrease, since multiplying by a fraction between 0 and 1 makes numbers smaller.
Understanding the common ratio is essential because it dictates the behavior and pattern of the entire sequence.
Sum of Terms
The sum of the terms in a geometric sequence can be calculated using a specialized formula, which is helpful when dealing with multiple terms. For the first \( n \) terms, the formula is \( S_n = a_1 \frac{1-r^n}{1-r} \).

In the solution provided, we find the sum of the first 12 terms using this formula. With \( a_1 = 59049 \) and \( r = -\frac{1}{3} \), it becomes important to calculate \( r^{12} \) carefully. This specific calculation showed how quickly powers can make very small numbers, which led to \( r^{12} = \frac{1}{531441} \).

Always remember that the sum relies heavily on correctly computing these powers and managing the fractions appropriately. This formula is extremely useful when you need to find the sum of a lot of terms without manually adding each one.
Sequence Formula
The basic formula to find any term in a geometric sequence is \( a_n = a_1 \, r^{n-1} \). This formula is critical because it allows us to calculate any specific term without computing all the previous ones.

In our task, finding terms such as the 33rd term means using these specific values: \( a_1 = 59049 \) and \( r = -\frac{1}{3} \). Thus, \( a_{33} = 59049 \, (-\frac{1}{3})^{32} \). It's crucial to note the exponent here. Since there were many terms until the 33rd, this process exploits the power of exponents to streamline calculations.

This formula encapsulates the essence of a geometric sequence, emphasizing both the exponential nature through \( r^{n-1} \) and its dependence on the very first term.
Geometric Series
A geometric series is simply the sum of the terms in a geometric sequence. In a broader context, it involves identifying how terms accumulate or grow when summed together. This concept is very different from just identifying individual sequence terms.

An interesting aspect of a geometric series, as we saw in the task, is that even when individual terms might become very small, their series adds significant value. The formula \( S_n = a_1 \frac{1-r^n}{1-r} \) is vital here, as it simplifies the process of summing potentially large numbers of terms.

Note how changing the common ratio can drastically influence the sum. In particular, when \( |r| < 1 \), it causes the series to converge, meaning the sum levels off. This property is key in many fields of mathematics and applications.

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

Consider the sequence defined by \(a_{1}=1, a_{n+1}=\) \((n+1)^{2}-a_{n}\) for \(n \geq 1 .\) Find the first six terms. Guess a general formula for \(a_{n}\) and prove that your answer is correct.

Give recursive definitions of each of the following sequences: (a) \([\mathrm{BB}] 1,5,5^{2}, 5^{3}, 5^{4}, \ldots\) (b) \(5,3,1,-1,-3, \ldots\) (c) \(4,1,3,-2,5,-7,12,-19,31, \ldots\) (d) \(1,2,0,3,-1,4,-2, \ldots\)

Define the sequence \(a_{1}, a_{2}, a_{3}, \ldots\) by \(a_{1}=0, a_{2}=\frac{1}{2}\) and \(a_{k+2}=\frac{1}{2}\left(a_{k}+a_{k+1}\right)\) for \(k \geq 1 .\) Find the first seven terms of this sequence. Prove that \(a_{n}=\frac{1}{3}\left(1-\left(-\frac{1}{2}\right)^{n-1}\right)\) for every \(n \geq 1\)

One of several differences between the Canadian and American games of football is that in Canada, a team can score a single point without first having scored a touchdown. So it is clear that any score is possible in the Canadian game. Is this so in the American game? Indeed this seems to be the case, even assuming (this is not true!) that in the United States, points can be scored only three at a time (with a field goal) or seven at a time (with a converted touchdown). Here is an argument. Assume that \(k\) points can be achieved with multiples of 3 or 7 . Here's how to reach \(k+1\) points. If \(k\) points are achieved with at least two field goals, subtracting these and adding a touchdown gives \(k+1\) points. On the other hand, if the \(k\) points are achieved with at least two touchdowns, subtracting these and adding five field goals also gives \(k+1\) points. Does this argument show that any score is possible in American football? Can it be used to show something about the nature of possible scores?

Let \(a_{1}, a_{2}, a_{3}, \ldots\) be the sequence defined by \(a_{1}=1\), \(a_{k+1}=3 a_{k}\) for \(k \geq 1\). Prove that \(a_{n}=3^{n-1}\) for all \(n \geq 1\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

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.