The first two terms of arithmetic sequences are shown in each case below. Determine the value of the requested term. Show the work that leads to your answer. \( \begin{array}{ll}\text { a) } a_{1}=7 \text { and } a_{2}=20 \text {, find } a_{2 s} & \text { (b) } b_{1}=9 \text { and } b_{2}=4, \text { find } b_{30}\end{array} \) N-GEN MATH ALGEBRA I - UNIT 4 - LINEAR FUNCTIONS-LESSON 14 EMATHINstruction, RED HOOK, NY 12571, O2022

Certainly! Let's solve each part of the problem step by step. ### **a) Given \( a_1 = 7 \) and \( a_2 = 20 \), find \( a_{2s} \)** **Step 1: Determine the Common Difference (\( d \))** An arithmetic sequence has a constant difference between consecutive terms. \[ d = a_2 - a_1 = 20 - 7 = 13 \] **Step 2: Find the General Formula for the \( n \)-th Term** The general formula for the \( n \)-th term of an arithmetic sequence is: \[ a_n = a_1 + (n - 1)d \] Plugging in the known values: \[ a_n = 7 + (n - 1) \times 13 = 7 + 13n - 13 = 13n - 6 \] **Step 3: Compute \( a_{2s} \)** Assuming \( s \) is a specific term number or a variable representing an integer, we substitute \( n = 2s \): \[ a_{2s} = 13(2s) - 6 = 26s - 6 \] **Answer for Part a:** \[ a_{2s} = 26s - 6 \] --- ### **b) Given \( b_1 = 9 \) and \( b_2 = 4 \), find \( b_{30} \)** **Step 1: Determine the Common Difference (\( d \))** \[ d = b_2 - b_1 = 4 - 9 = -5 \] **Step 2: Find the General Formula for the \( n \)-th Term** \[ b_n = b_1 + (n - 1)d \] Plugging in the known values: \[ b_n = 9 + (n - 1)(-5) = 9 - 5(n - 1) = 9 - 5n + 5 = 14 - 5n \] **Step 3: Compute \( b_{30} \)** Substitute \( n = 30 \): \[ b_{30} = 14 - 5 \times 30 = 14 - 150 = -136 \] **Answer for Part b:** \[ b_{30} = -136 \]