Find the 12th term of the arithmetic sequence 9,7, 5, .. Answer: Find the first term and the common difference of the arithmetic sequence whose 10th term is 34 and I term is 43. First term is Common difference is
Find the 12th term of the arithmetic sequence 9,7, 5, .. Answer: Find the first term and the common difference of the arithmetic sequence whose 10th term is 34 and I term is 43. First term is Common difference is
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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![### Arithmetic Sequences - Sample Question
#### Example Problem:
Find the first term and the common difference of the arithmetic sequence whose 10th term is 34 and 7th term is 19.
#### Solution:
Given:
- 10th term (\(a_{10}\)) = 34
- 7th term (\(a_7\)) = 19
To determine the first term (\(a\)) and the common difference (\(d\)), follow these steps:
1. Arithmetic sequence formula:
\[ a_n = a + (n-1)d \]
2. Plug in the given values to set up equations:
\[ a + 9d = 34 \quad \text{(10th term)} \]
\[ a + 6d = 19 \quad \text{(7th term)} \]
3. Subtract the second equation from the first:
\[ (a + 9d) - (a + 6d) = 34 - 19 \]
\[ 3d = 15 \]
\[ d = 5 \]
4. Substitute \(d\) back into one of the original equations to find \(a\):
\[ a + 6(5) = 19 \]
\[ a + 30 = 19 \]
\[ a = 19 - 30 \]
\[ a = -11 \]
Therefore, the first term (\(a\)) is \(-11\) and the common difference (\(d\)) is \(5\).
**Answer:**
- First Term (\(a\)): \(-11\)
- Common Difference (\(d\)): \(5\)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7921e840-b4be-4997-af5c-a45fa0acee7f%2F9f0abfc1-d514-4380-8be8-ab3c47e3317e%2Fvy1pbcr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Arithmetic Sequences - Sample Question
#### Example Problem:
Find the first term and the common difference of the arithmetic sequence whose 10th term is 34 and 7th term is 19.
#### Solution:
Given:
- 10th term (\(a_{10}\)) = 34
- 7th term (\(a_7\)) = 19
To determine the first term (\(a\)) and the common difference (\(d\)), follow these steps:
1. Arithmetic sequence formula:
\[ a_n = a + (n-1)d \]
2. Plug in the given values to set up equations:
\[ a + 9d = 34 \quad \text{(10th term)} \]
\[ a + 6d = 19 \quad \text{(7th term)} \]
3. Subtract the second equation from the first:
\[ (a + 9d) - (a + 6d) = 34 - 19 \]
\[ 3d = 15 \]
\[ d = 5 \]
4. Substitute \(d\) back into one of the original equations to find \(a\):
\[ a + 6(5) = 19 \]
\[ a + 30 = 19 \]
\[ a = 19 - 30 \]
\[ a = -11 \]
Therefore, the first term (\(a\)) is \(-11\) and the common difference (\(d\)) is \(5\).
**Answer:**
- First Term (\(a\)): \(-11\)
- Common Difference (\(d\)): \(5\)
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