6 The sum of the first three terms in a geometric series is 31 , the ratio. next three terms is 3875 . Find the first term and the common rato 7 The sum to \( n \) terms of a geometric series is 728 . If the first term is 2 and the last term is 486 , find the common ratio and the number of terms. 8 Wouter wants to buy a mathematical dictionary to improve his vocabulary The book costs R1 020. He saves R4 in the first month, R8 in the second month and R16 in the third month. If he continues in this manner, how long will it take him to save the money to buy the book?

Let's tackle each of your problems one by one. --- ## **Problem 6** **Statement:** The sum of the first three terms in a geometric series is 31, and the sum of the next three terms is 3875. Find the first term and the common ratio. **Solution:** Let's denote: - First term by \( a \) - Common ratio by \( r \) **Given:** 1. Sum of the first three terms: \[ S_3 = a + ar + ar^2 = 31 \quad \text{(Equation 1)} \] 2. Sum of the next three terms (4th, 5th, and 6th terms): \[ S_{\text{next 3}} = ar^3 + ar^4 + ar^5 = 3875 \quad \text{(Equation 2)} \] **Approach:** Notice that \( S_{\text{next 3}} = ar^3 (1 + r + r^2) \). Therefore, Equation 2 can be written as: \[ ar^3 (1 + r + r^2) = 3875 \] From Equation 1: \[ a (1 + r + r^2) = 31 \] Divide Equation 2 by Equation 1: \[ \frac{ar^3 (1 + r + r^2)}{a (1 + r + r^2)} = \frac{3875}{31} \] \[ r^3 = 125 \] \[ r = \sqrt[3]{125} = 5 \] Now, substitute \( r = 5 \) back into Equation 1 to find \( a \): \[ a (1 + 5 + 25) = 31 \] \[ a \times 31 = 31 \] \[ a = 1 \] **Answer:** - **First term (\( a \)) = 1** - **Common ratio (\( r \)) = 5** --- ## **Problem 7** **Statement:** The sum to \( n \) terms of a geometric series is 728. If the first term is 2 and the last term is 486, find the common ratio and the number of terms. **Solution:** Let's denote: - First term by \( a = 2 \) - Common ratio by \( r \) - Number of terms by \( n \) - Last term by \( l = 486 \) **Given:** 1. Sum of \( n \) terms: \[ S_n = a \frac{r^n - 1}{r - 1} = 728 \quad \text{(Equation 1)} \] 2. Last term: \[ l = a r^{n-1} = 486 \quad \text{(Equation 2)} \] **Approach:** From Equation 2: \[ 2 r^{n-1} = 486 \] \[ r^{n-1} = 243 \] \[ 243 = 3^5 \quad \text{(since \( 3^5 = 243 \))} \] This suggests that \( r = 3 \) and \( n-1 = 5 \), hence \( n = 6 \). **Verification:** Check if these values satisfy Equation 1: \[ S_6 = 2 \frac{3^6 - 1}{3 - 1} = 2 \frac{729 - 1}{2} = 2 \times 364 = 728 \] It matches the given sum. **Answer:** - **Common ratio (\( r \)) = 3** - **Number of terms (\( n \)) = 6** --- ## **Problem 8** **Statement:** Wouter wants to buy a mathematical dictionary costing R1,020. He saves R4 in the first month, R8 in the second month, and R16 in the third month. If he continues in this manner, how long will it take him to save the money to buy the book? **Solution:** Let's analyze the saving pattern: - **First month:** R4 - **Second month:** R8 - **Third month:** R16 - **...and so on.** This is a **geometric series** where: - **First term (\( a \)) = R4** - **Common ratio (\( r \)) = 2** (since each month's saving is double the previous month's saving) **Objective:** Find the number of months (\( n \)) required for the total savings to reach at least R1,020. **Sum of \( n \) terms in a geometric series:** \[ S_n = a \frac{r^n - 1}{r - 1} \] \[ S_n = 4 \times \frac{2^n - 1}{2 - 1} = 4 (2^n - 1) \] We need \( S_n \geq 1020 \): \[ 4 (2^n - 1) \geq 1020 \] \[ 2^n - 1 \geq \frac{1020}{4} = 255 \] \[ 2^n \geq 256 \] \[ 256 = 2^8 \] \[ \Rightarrow n \geq 8 \] **Verification:** Total savings after 8 months: \[ S_8 = 4 (2^8 - 1) = 4 (256 - 1) = 4 \times 255 = 1020 \text{ R} \] **Answer:** - **Wouter needs to save for 8 months to afford the dictionary.** ---