Question

3. The first three terms of the sequence \( 6 ; x ; y ; 27 \) form an arithmetic progression and the last three terms form a geometric progression. 3.1 Determine the values of \( x \) and \( y \). 4.2 State the arithmetic and geometric sequences for each of your solutions.

Ask by Conner Burton. in South Africa
Jan 22,2025

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Answer

**Solutions:** 1. **First Solution**: - \( x = 12 \), \( y = 18 \) - **Arithmetic Sequence**: \( 6, 12, 18 \) with a common difference of \( 6 \) - **Geometric Sequence**: \( 12, 18, 27 \) with a common ratio of \( 1.5 \) 2. **Second Solution**: - \( x = 0.75 \), \( y = -4.5 \) - **Arithmetic Sequence**: \( 6, 0.75, -4.5 \) with a common difference of \( -5.25 \) - **Geometric Sequence**: \( 0.75, -4.5, 27 \) with a common ratio of \( -6 \)

Solution

### Problem 3.1: Determine the values of \( x \) and \( y \). Given the sequence \( 6 ; x ; y ; 27 \): 1. **Arithmetic Progression (AP) for the first three terms**: For an arithmetic progression, the difference between consecutive terms is constant. Therefore: \[ x - 6 = y - x \implies 2x = 6 + y \implies y = 2x - 6 \] 2. **Geometric Progression (GP) for the last three terms**: For a geometric progression, the ratio between consecutive terms is constant. Therefore: \[ \frac{y}{x} = \frac{27}{y} \implies y^2 = 27x \] 3. **Substituting \( y = 2x - 6 \) into the GP equation**: \[ (2x - 6)^2 = 27x \] \[ 4x^2 - 24x + 36 = 27x \] \[ 4x^2 - 51x + 36 = 0 \] 4. **Solving the quadratic equation**: \[ x = \frac{51 \pm \sqrt{51^2 - 4 \cdot 4 \cdot 36}}{2 \cdot 4} = \frac{51 \pm \sqrt{2025}}{8} = \frac{51 \pm 45}{8} \] This gives two solutions: - \( x = \frac{51 + 45}{8} = \frac{96}{8} = 12 \) - \( x = \frac{51 - 45}{8} = \frac{6}{8} = 0.75 \) 5. **Finding corresponding \( y \) values**: - If \( x = 12 \): \[ y = 2(12) - 6 = 24 - 6 = 18 \] - If \( x = 0.75 \): \[ y = 2(0.75) - 6 = 1.5 - 6 = -4.5 \] **Solutions**: - \( x = 12 \) and \( y = 18 \) - \( x = 0.75 \) and \( y = -4.5 \) --- ### Problem 4.2: State the arithmetic and geometric sequences for each of your solutions. **1. When \( x = 12 \) and \( y = 18 \):** - **Arithmetic Progression (AP)**: \[ 6 ;\ 12 ;\ 18 \] - Common difference \( d = 6 \) (since \( 12 - 6 = 6 \) and \( 18 - 12 = 6 \)). - **Geometric Progression (GP)**: \[ 12 ;\ 18 ;\ 27 \] - Common ratio \( r = 1.5 \) (since \( \frac{18}{12} = 1.5 \) and \( \frac{27}{18} = 1.5 \)). **2. When \( x = 0.75 \) and \( y = -4.5 \):** - **Arithmetic Progression (AP)**: \[ 6 ;\ 0.75 ;\ -4.5 \] - Common difference \( d = -5.25 \) (since \( 0.75 - 6 = -5.25 \) and \( -4.5 - 0.75 = -5.25 \)). - **Geometric Progression (GP)**: \[ 0.75 ;\ -4.5 ;\ 27 \] - Common ratio \( r = -6 \) (since \( \frac{-4.5}{0.75} = -6 \) and \( \frac{27}{-4.5} = -6 \)). --- **Summary of Solutions**: 1. **First Solution**: - \( x = 12 \), \( y = 18 \) - **AP**: \( 6 ;\ 12 ;\ 18 \) with common difference \( 6 \) - **GP**: \( 12 ;\ 18 ;\ 27 \) with common ratio \( 1.5 \) 2. **Second Solution**: - \( x = 0.75 \), \( y = -4.5 \) - **AP**: \( 6 ;\ 0.75 ;\ -4.5 \) with common difference \( -5.25 \) - **GP**: \( 0.75 ;\ -4.5 ;\ 27 \) with common ratio \( -6 \)

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To solve for \( x \) and \( y \) in the sequence \( 6; x; y; 27 \), we start with the conditions for arithmetic and geometric progressions. For the arithmetic progression \( 6, x, y, 27 \): The difference between consecutive terms is constant, so we have: \( x - 6 = y - x = 27 - y \). Let the common difference be \( d \). Then we can express \( x \) and \( y \) as: \( x = 6 + d \) and \( y = 6 + 2d \). Setting \( 27 - y = d \) gives: \( 27 - (6 + 2d) = d \) This leads to: \( 27 - 6 - 2d = d \) or \( 21 = 3d \), giving \( d = 7 \). Now substituting \( d = 7 \): \( x = 6 + 7 = 13 \) and \( y = 6 + 2(7) = 20 \). Next, for the geometric progression sequence \( x, y, 27 \): Here, the ratios between consecutive terms are equal: \(\frac{y}{x} = \frac{27}{y}\). Substituting \( x = 13 \) and \( y = 20 \) gives: \(\frac{20}{13} = \frac{27}{20}\) which is indeed true. Now, we state the sequences: Arithmetic: \( 6, 13, 20, 27 \) with a common difference of \( 7 \). Geometric: \( 13, 20, 27 \) with a common ratio \( \frac{20}{13} \) or approximately \( 1.538 \).

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