EXERCISE 5 1. Determine the sum of each geometric series (use an appropriate formula): (a) $$ \frac{2}{27}+\frac{2}{9}+\frac{2}{3}+\ldots $$. to 9 terms (b) $$ -64+32-16+\ldots $$. to 10 terms

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**Exercise 5**

**1. Determine the sum of each geometric series (use an appropriate formula):**

A geometric series is a series of the form:
$$
a_1 + a_1 r + a_1 r^2 + \ldots + a_1 r^{n-1}
$$
where:
- $$ a_1 $$ is the first term,
- $$ r $$ is the common ratio,
- $$ n $$ is the number of terms.

The sum of the first $$ n $$ terms of a geometric series is given by:
$$
S_n = \frac{a_1 (1 - r^n)}{1 - r}, \quad \text{provided } r \neq 1
$$

Let's apply this formula to each part of the exercise.

---

### **1(a).** $$ \frac{2}{27} + \frac{2}{9} + \frac{2}{3} + \ldots $$ (to 9 terms)

**Step 1: Identify the components of the geometric series.**

- **First term ($$ a_1 $$)**: $$ \frac{2}{27} $$
- **Common ratio ($$ r $$)**: To find $$ r $$, divide the second term by the first term:
  $$
  r = \frac{\frac{2}{9}}{\frac{2}{27}} = \frac{2}{9} \times \frac{27}{2} = 3
  $$
- **Number of terms ($$ n $$)**: 9

**Step 2: Apply the sum formula.**

$$
S_9 = \frac{\frac{2}{27} (1 - 3^9)}{1 - 3}
$$

**Step 3: Calculate $$ 3^9 $$.**

$$
3^9 = 19,683
$$

**Step 4: Plug the values into the formula.**

$$
S_9 = \frac{\frac{2}{27} (1 - 19,683)}{1 - 3} = \frac{\frac{2}{27} (-19,682)}{-2}
$$

**Step 5: Simplify the expression.**

$$
S_9 = \frac{2 \times (-19,682)}{27 \times (-2)} = \frac{-39,364}{-54} = \frac{39,364}{54}
$$

**Step 6: Reduce the fraction (if possible).**

$$
\frac{39,364}{54} = \frac{19,682}{27} \approx 729.7037
$$

**Final Answer for 1(a):**

$$
S_9 = \frac{19,682}{27} \quad \text{or approximately } 729.70
$$

---

### **1(b).** $$ -64 + 32 - 16 + \ldots $$ (to 10 terms)

**Step 1: Identify the components of the geometric series.**

- **First term ($$ a_1 $$)**: $$ -64 $$
- **Common ratio ($$ r $$)**: To find $$ r $$, divide the second term by the first term:
  $$
  r = \frac{32}{-64} = -0.5
  $$
- **Number of terms ($$ n $$)**: 10

**Step 2: Apply the sum formula.**

$$
S_{10} = \frac{-64 (1 - (-0.5)^{10})}{1 - (-0.5)}
$$

**Step 3: Calculate $$ (-0.5)^{10} $$.**

$$
(-0.5)^{10} = (0.5)^{10} = \frac{1}{1024}
$$

**Step 4: Plug the values into the formula.**

$$
S_{10} = \frac{-64 \left(1 - \frac{1}{1024}\right)}{1 + 0.5} = \frac{-64 \left(\frac{1023}{1024}\right)}{1.5}
$$

**Step 5: Simplify the expression.**

$$
S_{10} = \frac{-64 \times 1023}{1024 \times 1.5} = \frac{-65,472}{1,536} = -42.6 \quad (\text{rounded to one decimal place})
$$

**Alternatively, simplifying the fractions:**

$$
\frac{-64 \times 1023}{1024 \times 1.5} = \frac{-64 \times 1023}{1,536} = \frac{-64}{1,536} \times 1023 = \frac{-1}{24} \times 1023 = -42.625
$$

**Final Answer for 1(b):**

$$
S_{10} = -42.625
$$

---

**Summary of Results:**

- **1(a)**: The sum of the first 9 terms is $$ \frac{19,682}{27} $$ or approximately **729.70**.
- **1(b)**: The sum of the first 10 terms is **-42.625**.
**Exercise 5 Solutions:** 1. **(a)** The sum of the first 9 terms is $$ \frac{19,682}{27} $$ or approximately **729.70**. 2. **(b)** The sum of the first 10 terms is **-42.625**.

EXERCISE 5

Determine the sum of each geometric series (use an appropriate formula):
(a) . to 9 terms
(b) . to 10 terms

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Solution

1(a). (to 9 terms)

1(b). (to 10 terms)

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EXERCISE 5 Determine the sum of each geometric series (use an appropriate formula): (a) . to 9 terms (b) . to 10 terms