Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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**Problem Statement:** Find the sum of the first 8 terms for the following arithmetic sequence. Given: - \( a_2 = 9 \) - \( a_4 = 19 \) **Question:** The sum of the first 8 terms is [ ]. **Explanation:** To solve this problem, follow these steps: 1. **Identify the common difference**: - Use the formula of the n-th term of an arithmetic sequence, which is \( a_n = a_1 + (n-1)d \). - Here, \( a_2 = 9 \) and \( a_4 = 19 \). - Set up the equations: - \( a_2 = a_1 + d = 9 \) - \( a_4 = a_1 + 3d = 19 \) - Solve these equations simultaneously to find \( a_1 \) and \( d \). 2. **Calculate the sum of the first 8 terms**: - The formula for the sum of the first n terms of an arithmetic sequence is: \[ S_n = \frac{n}{2} \times (2a_1 + (n-1)d) \] - Substitute \( n = 8 \), along with the values for \( a_1 \) and \( d \), to find \( S_8 \). This approach will yield the required sum for the first 8 terms of the sequence.
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Transcribed Image Text:**Problem Statement:** Find the sum of the first 8 terms for the following arithmetic sequence. Given: - \( a_2 = 9 \) - \( a_4 = 19 \) **Question:** The sum of the first 8 terms is [ ]. **Explanation:** To solve this problem, follow these steps: 1. **Identify the common difference**: - Use the formula of the n-th term of an arithmetic sequence, which is \( a_n = a_1 + (n-1)d \). - Here, \( a_2 = 9 \) and \( a_4 = 19 \). - Set up the equations: - \( a_2 = a_1 + d = 9 \) - \( a_4 = a_1 + 3d = 19 \) - Solve these equations simultaneously to find \( a_1 \) and \( d \). 2. **Calculate the sum of the first 8 terms**: - The formula for the sum of the first n terms of an arithmetic sequence is: \[ S_n = \frac{n}{2} \times (2a_1 + (n-1)d) \] - Substitute \( n = 8 \), along with the values for \( a_1 \) and \( d \), to find \( S_8 \). This approach will yield the required sum for the first 8 terms of the sequence.
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