/*! 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 31 Find the ninth term of a geometr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 31

Find the ninth term of a geometric sequence whose fourth term is 4 and whose seventh term is 5 .

Short Answer

Expert verified
The ninth term of the geometric sequence is approximately 6.4.

Step by step solution

01

Identify the formula for a geometric sequence

A geometric sequence is a sequence of numbers in which each term after the first one is a constant multiple of the previous term. This constant multiple is known as the common ratio, denoted as r. The general formula for a geometric sequence is \(a_n = a_1 \cdot r^{(n-1)}\), where \(a_n\) denotes the nth term, \(a_1\) is the first term, and n is the number of terms.
02

Write the formula for given terms

We are given that the fourth term (\(a_4\)) is 4 and the seventh term (\(a_7\)) is 5. Let's write the formula for these terms: \(a_4 = a_1 \cdot r^{(4-1)} = a_1 \cdot r^3\) \(a_7 = a_1 \cdot r^{(7-1)} = a_1 \cdot r^6\)
03

Find the common ratio (r)

We can compute the common ratio by dividing the equation for the seventh term by the equation for the fourth term: \(\frac{a_7}{a_4} = \frac{a_1 \cdot r^6}{a_1 \cdot r^3}\) Now, substitute the given values for \(a_4\) and \(a_7\): \(\frac{5}{4} = \frac{a_1 \cdot r^6}{a_1 \cdot r^3}\) As we can see, \(a_1\) will cancel out on both sides, leaving us with: \(\frac{5}{4} = \frac{r^6}{r^3}\) Simplify the equation to find the value of r: \(r^3 = \frac{5}{4}\) Now take the cube root of both sides: \(r = \sqrt[3]{\frac{5}{4}}\)
04

Find the first term (a_1)

Now that we have determined the common ratio, we can find the first term (\(a_1\)). Using the equation from step 2, let's plug in the values of \(a_4\) and r: \(4 = a_1 \cdot \left(\sqrt[3]{\frac{5}{4}}\right)^3\) Solve for \(a_1\): \(a_1 = \frac{4}{\left(\sqrt[3]{\frac{5}{4}}\right)^3} = \frac{4}{\frac{5}{4}}\) Therefore, \(a_1 = 3.2\)
05

Find the ninth term (a_9)

Now that we have the first term and the common ratio, we can find the ninth term (\(a_9\)) using the general formula: \(a_9 = a_1 \cdot r^{(9-1)} = 3.2 \cdot \left(\sqrt[3]{\frac{5}{4}}\right)^8\) After calculating, we get: \(a_9 \approx 6.4\) So, the ninth term of the geometric sequence is approximately 6.4.

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!

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 15-24, evaluate the geometric series. \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\cdots+\frac{1}{3^{33}}\)

Evaluate \(\lim _{n \rightarrow \infty} n\left(\ln \left(7+\frac{1}{n}\right)-\ln 7\right)\)

Define a recursive sequence by \(a_{1}=3 \quad\) and \(\quad a_{n+1}=\frac{1}{2}\left(\frac{7}{a_{n}}+a_{n}\right)\) for \(n \geq 1 .\) Find the smallest value of \(n\) such that \(a_{n}\) agrees with \(\sqrt{7}\) for at least six digits after the decimal point.

Find the tenth term of a geometric sequence whose second term is 3 and whose seventh term is \(11 .\)

Learn about Zeno's paradox (from a book, a friend, or a web search) and then relate the explanation of this ancient Greek problem to the infinite series $$ \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots=1. $$
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

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.