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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![### Summing an Arithmetic Series
#### Problem Statement:
Use a formula for arithmetic series to find the sum of the series:
\[ 1 + 3 + 5 + 7 + \ldots + 197 + 199 \]
#### Explanation:
This problem involves finding the sum of an arithmetic series.
An arithmetic series is the sum of the terms in an arithmetic sequence. An arithmetic sequence is a sequence of numbers in which the difference between any two consecutive terms is always the same. This difference is known as the common difference \(d\).
In our given series:
\[ 1, 3, 5, 7, \ldots, 197, 199 \]
The first term \(a = 1\) and the common difference \(d = 2\).
To find the sum of the first \(n\) terms of an arithmetic series, we use the formula:
\[ S_n = \frac{n}{2} [2a + (n-1)d] \]
where:
- \(S_n\) is the sum of the first \(n\) terms,
- \(a\) is the first term,
- \(d\) is the common difference,
- \(n\) is the number of terms.
#### Steps to Find the Sum:
1. **Identify the first term \(a\), and the common difference \(d\)**:
- First term \(a = 1\)
- Common difference \(d = 2\)
2. **Find the number of terms \(n\)**:
The last term \(l\) of the arithmetic series is 199.
Using the nth-term formula of arithmetic sequence:
\[ l = a + (n-1)d \]
Solve for \(n\):
\[ 199 = 1 + (n-1) \cdot 2 \]
\[ 199 = 1 + 2n - 2 \]
\[ 199 = 2n - 1 \]
\[ 200 = 2n \]
\[ n = 100 \]
So, there are 100 terms in the series.
3. **Calculate the sum \(S_n\) using the sum formula**:
\[ S_{100} = \frac{100}{2} [2 \cdot 1 + (100-1)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9e61a896-eac2-4e92-aac1-4d7f245a1e73%2Ffb107186-3267-47a5-8e90-683234e4a571%2F1vjv7vx.jpeg&w=3840&q=75)
Transcribed Image Text:### Summing an Arithmetic Series
#### Problem Statement:
Use a formula for arithmetic series to find the sum of the series:
\[ 1 + 3 + 5 + 7 + \ldots + 197 + 199 \]
#### Explanation:
This problem involves finding the sum of an arithmetic series.
An arithmetic series is the sum of the terms in an arithmetic sequence. An arithmetic sequence is a sequence of numbers in which the difference between any two consecutive terms is always the same. This difference is known as the common difference \(d\).
In our given series:
\[ 1, 3, 5, 7, \ldots, 197, 199 \]
The first term \(a = 1\) and the common difference \(d = 2\).
To find the sum of the first \(n\) terms of an arithmetic series, we use the formula:
\[ S_n = \frac{n}{2} [2a + (n-1)d] \]
where:
- \(S_n\) is the sum of the first \(n\) terms,
- \(a\) is the first term,
- \(d\) is the common difference,
- \(n\) is the number of terms.
#### Steps to Find the Sum:
1. **Identify the first term \(a\), and the common difference \(d\)**:
- First term \(a = 1\)
- Common difference \(d = 2\)
2. **Find the number of terms \(n\)**:
The last term \(l\) of the arithmetic series is 199.
Using the nth-term formula of arithmetic sequence:
\[ l = a + (n-1)d \]
Solve for \(n\):
\[ 199 = 1 + (n-1) \cdot 2 \]
\[ 199 = 1 + 2n - 2 \]
\[ 199 = 2n - 1 \]
\[ 200 = 2n \]
\[ n = 100 \]
So, there are 100 terms in the series.
3. **Calculate the sum \(S_n\) using the sum formula**:
\[ S_{100} = \frac{100}{2} [2 \cdot 1 + (100-1)
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