2 The first term of a geometric series is and the 6 th term is . Determine:
a the common ratio
b the sum of the first 10 terms, conect to one decimal place.
3 The first term of a geometric series is 54 and the common ratio is .
Calculate the value of that will give a sum of 80 .

Ask by Vargas Clark. in South Africa

Jan 23,2025

Question

2 The first term of a geometric series is and the 6 th term is . Determine:
a the common ratio
b the sum of the first 10 terms, conect to one decimal place.
3 The first term of a geometric series is 54 and the common ratio is .
Calculate the value of that will give a sum of 80 .

Ask by Vargas Clark. in South Africa

Jan 23,2025

UpStudy AI · Accepted Answer

Let's solve each part of your questions step-by-step.

---

## **Question 2**

**Given:**
- First term ($$ a_1 $$) = $$ \frac{3}{128} $$
- 6th term ($$ a_6 $$) = $$ \frac{3}{4} $$

### **a) Determine the Common Ratio ($$ r $$)**

The $$ n $$-th term of a geometric series is given by:
$$
a_n = a_1 \times r^{n-1}
$$

For the 6th term:
$$
a_6 = a_1 \times r^{5}
$$
$$
\frac{3}{4} = \frac{3}{128} \times r^{5}
$$

**Solving for $$ r $$:**
$$
\frac{3}{4} = \frac{3}{128} \times r^{5}
$$
$$
\frac{3}{4} \div \frac{3}{128} = r^{5}
$$
$$
\frac{3}{4} \times \frac{128}{3} = r^{5}
$$
$$
\frac{128}{4} = r^{5}
$$
$$
32 = r^{5}
$$

Since $$ 32 = 2^5 $$, we have:
$$
r = 2
$$

**Answer:**  
The common ratio $$ r $$ is **2**.

---

### **b) Determine the Sum of the First 10 Terms ($$ S_{10} $$)**

The sum of the first $$ n $$ terms in a geometric series is given by:
$$
S_n = a_1 \times \frac{r^{n} - 1}{r - 1}
$$

**Given:**
- $$ a_1 = \frac{3}{128} $$
- $$ r = 2 $$
- $$ n = 10 $$

**Calculating $$ S_{10} $$:**
$$
S_{10} = \frac{3}{128} \times \frac{2^{10} - 1}{2 - 1}
$$
$$
S_{10} = \frac{3}{128} \times \frac{1024 - 1}{1}
$$
$$
S_{10} = \frac{3}{128} \times 1023
$$
$$
S_{10} = \frac{3069}{128} \approx 24.0 \quad (\text{to one decimal place})
$$

**Answer:**  
The sum of the first 10 terms $$ S_{10} $$ is **24.0**.

---

## **Question 3**

**Given:**
- First term ($$ a_1 $$) = 54
- Common ratio ($$ r $$) = $$ \frac{1}{3} $$
- Desired sum ($$ S_n $$) = 80

**Find:**  
The value of $$ n $$ such that $$ S_n = 80 $$.

**Using the sum formula:**
$$
S_n = a_1 \times \frac{1 - r^{n}}{1 - r}
$$
$$
80 = 54 \times \frac{1 - \left(\frac{1}{3}\right)^{n}}{1 - \frac{1}{3}}
$$
$$
80 = 54 \times \frac{1 - \left(\frac{1}{3}\right)^{n}}{\frac{2}{3}}
$$
$$
80 = 54 \times \frac{3}{2} \times \left(1 - \left(\frac{1}{3}\right)^{n}\right)
$$
$$
80 = 81 \times \left(1 - \left(\frac{1}{3}\right)^{n}\right)
$$
$$
\frac{80}{81} = 1 - \left(\frac{1}{3}\right)^{n}
$$
$$
\left(\frac{1}{3}\right)^{n} = 1 - \frac{80}{81}
$$
$$
\left(\frac{1}{3}\right)^{n} = \frac{1}{81}
$$
$$
\left(\frac{1}{3}\right)^{n} = \left(\frac{1}{3}\right)^{4}
$$

**Therefore:**
$$
n = 4
$$

**Answer:**  
The value of $$ n $$ is **4**.
**Question 2:** - **a) Common Ratio:** 2 - **b) Sum of First 10 Terms:** 24.0 **Question 3:** - **Value of $$ n $$:** 4

2 The first term of a geometric series is and the 6 th term is . Determine:
a the common ratio
b the sum of the first 10 terms, conect to one decimal place.
3 The first term of a geometric series is 54 and the common ratio is .
Calculate the value of that will give a sum of 80 .

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2 The first term of a geometric series is and the 6 th term is . Determine: a the common ratio b the sum of the first 10 terms, conect to one decimal place. 3 The first term of a geometric series is 54 and the common ratio is . Calculate the value of that will give a sum of 80 .

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2 The first term of a geometric series is and the 6 th term is . Determine:
a the common ratio
b the sum of the first 10 terms, conect to one decimal place.
3 The first term of a geometric series is 54 and the common ratio is .
Calculate the value of that will give a sum of 80 .