3.2 The sum of an infinite, geometric series is \( 13,5(r \neq 1) \). The sum of the same series, calculated from the third term is 1,5 . 3.2.1 Calculate \( r \) if \( r>0 \). 3.2.2 Hence, determine the first THREE terms of the series.

To solve the given problem, let's break it down step by step. ### Given: 1. The sum of an infinite geometric series is **13.5**, with the common ratio \( r \neq 1 \). 2. The sum of the same series starting from the **third term** is **1.5**. ### Definitions: - **First term of the series**: \( a \) - **Common ratio**: \( r \) (where \( |r| < 1 \) for convergence) ### 3.2.1 Calculate \( r \) if \( r > 0 \) **Step 1: Express the sum of the infinite series.** The sum \( S \) of an infinite geometric series is given by: \[ S = \frac{a}{1 - r} \] Given \( S = 13.5 \), we have: \[ \frac{a}{1 - r} = 13.5 \quad \text{(Equation 1)} \] **Step 2: Express the sum starting from the third term.** The third term of the series is \( ar^2 \). The sum starting from the third term is: \[ S_{\text{from 3rd}} = ar^2 + ar^3 + ar^4 + \dots = \frac{ar^2}{1 - r} = 1.5 \quad \text{(Equation 2)} \] **Step 3: Divide Equation 2 by Equation 1 to eliminate \( a \).** \[ \frac{\frac{ar^2}{1 - r}}{\frac{a}{1 - r}} = r^2 = \frac{1.5}{13.5} = \frac{1}{9} \] \[ r^2 = \frac{1}{9} \implies r = \frac{1}{3} \quad (\text{since } r > 0) \] **Answer to 3.2.1:** \[ r = \frac{1}{3} \] ### 3.2.2 Determine the first THREE terms of the series. **Step 1: Find the first term \( a \) using Equation 1.** \[ \frac{a}{1 - \frac{1}{3}} = 13.5 \implies \frac{a}{\frac{2}{3}} = 13.5 \implies a = 13.5 \times \frac{2}{3} = 9 \] **Step 2: Calculate the first three terms using \( a \) and \( r \).** \[ \text{First term} = a = 9 \] \[ \text{Second term} = ar = 9 \times \frac{1}{3} = 3 \] \[ \text{Third term} = ar^2 = 9 \times \left(\frac{1}{3}\right)^2 = 1 \] **Answer to 3.2.2:** \[ \text{First three terms are } 9,\ 3,\ \text{and } 1. \]