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For the following geometric series, please find the first term "a" and the common ration "r": 2 2 2+ + 3 9 - 2 + + 27

For the following geometric series, please find the first term "a" and the common ration "r": 2 2 2+ + 3 9 - 2 + + 27

Calculus: Early Transcendentals
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...
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**Text:** For the following geometric series, please find the first term "a" and the common ratio "r": \[ 2 + \frac{2}{3} + \frac{2}{9} + \frac{2}{27} + \ldots \] **Explanation:** This is a geometric series where each term after the first is the product of the previous term and the common ratio \( r \). The first term \( a \) is 2. The series terms are as follows: - First term: \( 2 \) - Second term: \( \frac{2}{3} \) - Third term: \( \frac{2}{9} \) - Fourth term: \( \frac{2}{27} \) To find the common ratio \( r \), divide the second term by the first term: \[ r = \frac{\frac{2}{3}}{2} = \frac{1}{3} \] Each term is a factor of \( \frac{1}{3} \) of the previous term, confirming the common ratio \( r = \frac{1}{3} \).
Transcribed Image Text:**Text:** For the following geometric series, please find the first term "a" and the common ratio "r": \[ 2 + \frac{2}{3} + \frac{2}{9} + \frac{2}{27} + \ldots \] **Explanation:** This is a geometric series where each term after the first is the product of the previous term and the common ratio \( r \). The first term \( a \) is 2. The series terms are as follows: - First term: \( 2 \) - Second term: \( \frac{2}{3} \) - Third term: \( \frac{2}{9} \) - Fourth term: \( \frac{2}{27} \) To find the common ratio \( r \), divide the second term by the first term: \[ r = \frac{\frac{2}{3}}{2} = \frac{1}{3} \] Each term is a factor of \( \frac{1}{3} \) of the previous term, confirming the common ratio \( r = \frac{1}{3} \).
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