Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Question
![The image presents a mathematical expression for a sequence and poses a question about it:
The formula for the sequence is:
\[ c(n) = -6 \left( \frac{1}{3} \right)^{n-1} \]
The question asks:
"What is the 2nd term in the sequence?"
To find the 2nd term, substitute \( n = 2 \) into the formula:
\[ c(2) = -6 \left( \frac{1}{3} \right)^{2-1} = -6 \left( \frac{1}{3} \right)^1 = -6 \times \frac{1}{3} = -2 \]
Therefore, the 2nd term in the sequence is \(-2\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F330a48d4-f957-4611-80aa-7f79b718bbcf%2F828ac881-531d-4d59-b9aa-cd1125f04e49%2Fvhgty8r_processed.png&w=3840&q=75)
Transcribed Image Text:The image presents a mathematical expression for a sequence and poses a question about it:
The formula for the sequence is:
\[ c(n) = -6 \left( \frac{1}{3} \right)^{n-1} \]
The question asks:
"What is the 2nd term in the sequence?"
To find the 2nd term, substitute \( n = 2 \) into the formula:
\[ c(2) = -6 \left( \frac{1}{3} \right)^{2-1} = -6 \left( \frac{1}{3} \right)^1 = -6 \times \frac{1}{3} = -2 \]
Therefore, the 2nd term in the sequence is \(-2\).
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