3.2 The sum of an infinite, geometric series is $$ 13,5(r
eq 1) $$. The sum of the same series, calculated from the third term is 1,5 . 3.2.1 Calculate $$ r $$ if $$ r0 $$. 3.2.2 Hence, determine the first THREE terms of the series.

Question

3.2 The sum of an infinite, geometric series is $$ 13,5(r 
eq 1) $$. 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.

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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:**
$$
\text{First three terms are } 9,\ 3,\ \text{and } 1.
$$
$$ r = \frac{1}{3} $$ First three terms: 9, 3, 1

3.2 The sum of an infinite, geometric series is .
The sum of the same series, calculated from the third term is 1,5 .
3.2.1 Calculate if .
3.2.2 Hence, determine the first THREE terms of the series.

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3.2 The sum of an infinite, geometric series is . The sum of the same series, calculated from the third term is 1,5 . 3.2.1 Calculate if . 3.2.2 Hence, determine the first THREE terms of the series.