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Find the sum: 1+ 6+ 11 + ... + 41 Answer:

Find the sum: 1+ 6+ 11 + ... + 41 Answer:

Algebra and Trigonometry (6th Edition)
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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13.4: series and thier notations

find the sum

**Find the Sum of an Arithmetic Sequence** To solve the problem, we need to find the sum of the arithmetic sequence: \[ 1 + 6 + 11 + \ldots + 41 \] First, we identify the common difference in this sequence. The common difference \(d\) can be found by subtracting the first term from the second term: \[ d = 6 - 1 = 5 \] Now we need to find the number of terms \(n\) in this sequence. The general form of the \(n\)-th term of an arithmetic sequence is given by: \[ a_n = a + (n-1)d \] where \(a\) is the first term, \(d\) is the common difference, and \(a_n\) is the \(n\)-th term. We know that the last term of the sequence is 41, so we set up the equation: \[ 41 = 1 + (n-1) \cdot 5 \] Solving for \(n\): \[ 41 = 1 + 5(n-1) \] \[ 41 - 1 = 5(n-1) \] \[ 40 = 5(n-1) \] \[ 8 = n-1 \] \[ n = 9 \] There are 9 terms in this sequence. Now we use the formula for the sum \(S_n\) of the first \(n\) terms of an arithmetic sequence: \[ S_n = \frac{n}{2} \cdot (a + a_n) \] Plugging in the values we have: \[ S_9 = \frac{9}{2} \cdot (1 + 41) \] \[ S_9 = \frac{9}{2} \cdot 42 \] \[ S_9 = 9 \cdot 21 \] \[ S_9 = 189 \] **Answer:** \( 189 \) Highlighted Formula: \[ S_n = \frac{n}{2} \cdot (a + a_n) \] Explanation with Numerical Example: 1. Identify the common difference: \( d = 5 \) 2. Determine the number of terms: \( n = 9 \) 3. Find the sum: \( S_9 = 189 \)
Transcribed Image Text:**Find the Sum of an Arithmetic Sequence** To solve the problem, we need to find the sum of the arithmetic sequence: \[ 1 + 6 + 11 + \ldots + 41 \] First, we identify the common difference in this sequence. The common difference \(d\) can be found by subtracting the first term from the second term: \[ d = 6 - 1 = 5 \] Now we need to find the number of terms \(n\) in this sequence. The general form of the \(n\)-th term of an arithmetic sequence is given by: \[ a_n = a + (n-1)d \] where \(a\) is the first term, \(d\) is the common difference, and \(a_n\) is the \(n\)-th term. We know that the last term of the sequence is 41, so we set up the equation: \[ 41 = 1 + (n-1) \cdot 5 \] Solving for \(n\): \[ 41 = 1 + 5(n-1) \] \[ 41 - 1 = 5(n-1) \] \[ 40 = 5(n-1) \] \[ 8 = n-1 \] \[ n = 9 \] There are 9 terms in this sequence. Now we use the formula for the sum \(S_n\) of the first \(n\) terms of an arithmetic sequence: \[ S_n = \frac{n}{2} \cdot (a + a_n) \] Plugging in the values we have: \[ S_9 = \frac{9}{2} \cdot (1 + 41) \] \[ S_9 = \frac{9}{2} \cdot 42 \] \[ S_9 = 9 \cdot 21 \] \[ S_9 = 189 \] **Answer:** \( 189 \) Highlighted Formula: \[ S_n = \frac{n}{2} \cdot (a + a_n) \] Explanation with Numerical Example: 1. Identify the common difference: \( d = 5 \) 2. Determine the number of terms: \( n = 9 \) 3. Find the sum: \( S_9 = 189 \)
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