Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
Related questions
Question
![**Understanding Geometric Sequences**
**Question:**
What is the explicit rule for the nth term of the geometric sequence?
**Given Sequence:**
\[3, 18, 108, 648, 3,888, \ldots\]
**Answer Choices:**
1. \[ a_n = 3(6^n) \]
2. \[ a_n = 3(6^{n+1}) \]
3. \[ a_n = 6(3^{n-1}) \]
4. \[ a_n = 3(6^{n-1}) \]
**Explanation:**
A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous term by a constant called the common ratio.
To determine the common ratio (\(r\)), we divide any term by the previous term:
\[ r = \frac{18}{3} = 6 \]
\[ r = \frac{108}{18} = 6 \]
\[ r = \frac{648}{108} = 6 \]
Given that the common ratio is constant (6), we need to find a formula that properly represents this relationship.
To write the explicit formula for the nth term of a geometric sequence, we use the formula:
\[ a_n = a_1 \cdot r^{(n-1)} \]
Where:
- \(a_n\) = the nth term
- \(a_1\) = the first term of the sequence
- \(r\) = the common ratio
- \(n\) = the term number
Here:
- \(a_1 = 3\)
- \(r = 6\)
Plug these values into the formula:
\[ a_n = 3 \cdot 6^{(n-1)} \]
Thus, the correct choice is:
\[ \boxed{a_n = 3(6^{n-1})} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3047b756-b2f4-4f66-ba10-7c0773ac07fd%2Fb561995f-ac22-4fc2-9105-8f0f4a8807a4%2F8lkfnd8_processed.png&w=3840&q=75)
Transcribed Image Text:**Understanding Geometric Sequences**
**Question:**
What is the explicit rule for the nth term of the geometric sequence?
**Given Sequence:**
\[3, 18, 108, 648, 3,888, \ldots\]
**Answer Choices:**
1. \[ a_n = 3(6^n) \]
2. \[ a_n = 3(6^{n+1}) \]
3. \[ a_n = 6(3^{n-1}) \]
4. \[ a_n = 3(6^{n-1}) \]
**Explanation:**
A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous term by a constant called the common ratio.
To determine the common ratio (\(r\)), we divide any term by the previous term:
\[ r = \frac{18}{3} = 6 \]
\[ r = \frac{108}{18} = 6 \]
\[ r = \frac{648}{108} = 6 \]
Given that the common ratio is constant (6), we need to find a formula that properly represents this relationship.
To write the explicit formula for the nth term of a geometric sequence, we use the formula:
\[ a_n = a_1 \cdot r^{(n-1)} \]
Where:
- \(a_n\) = the nth term
- \(a_1\) = the first term of the sequence
- \(r\) = the common ratio
- \(n\) = the term number
Here:
- \(a_1 = 3\)
- \(r = 6\)
Plug these values into the formula:
\[ a_n = 3 \cdot 6^{(n-1)} \]
Thus, the correct choice is:
\[ \boxed{a_n = 3(6^{n-1})} \]
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning