Login
Premium
Home Question Bank Math Algebra
Question

4 The first three terms of the sequence \( 6 ; x ; y ; 16 \) form an arithmetic progression and the last three ferms form a geometric progression. 4.1 Determine \( x \) and \( y \). 4.2 State the arithmetic and geometric sequences for each of your solutions.

Ask by Barrett Mcfarlane. in South Africa
Jan 22,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

**4.1 Determine \( x \) and \( y \):** - **First Solution:** - \( x = 9 \) - \( y = 12 \) - **Second Solution:** - \( x = 1 \) - \( y = -4 \) **4.2 Sequences:** 1. **First Solution:** - **Arithmetic Sequence:** \( 6, 9, 12 \) (common difference: 3) - **Geometric Sequence:** \( 9, 12, 16 \) (common ratio: \( \frac{4}{3} \)) 2. **Second Solution:** - **Arithmetic Sequence:** \( 6, 1, -4 \) (common difference: -5) - **Geometric Sequence:** \( 1, -4, 16 \) (common ratio: -4)

Solution

**4.1 Determine \( x \) and \( y \).** We are given the sequence \( 6; x; y; 16 \) with the following conditions: 1. **Arithmetic Progression (AP)**: The first three terms \( 6, x, y \) form an AP. 2. **Geometric Progression (GP)**: The last three terms \( x, y, 16 \) form a GP. Let's translate these conditions into equations. **Arithmetic Progression Condition:** In an AP, the difference between consecutive terms is constant. Therefore: \[ x - 6 = y - x \implies 2x = y + 6 \implies y = 2x - 6 \] **Geometric Progression Condition:** In a GP, the ratio between consecutive terms is constant. Therefore: \[ \frac{y}{x} = \frac{16}{y} \implies y^2 = 16x \] Substituting \( y = 2x - 6 \) into the GP condition: \[ (2x - 6)^2 = 16x \\ 4x^2 - 24x + 36 = 16x \\ 4x^2 - 40x + 36 = 0 \\ x^2 - 10x + 9 = 0 \] Solving the quadratic equation \( x^2 - 10x + 9 = 0 \): \[ x = \frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm 8}{2} \] \[ x = 9 \quad \text{or} \quad x = 1 \] Using \( y = 2x - 6 \): - If \( x = 9 \): \[ y = 2(9) - 6 = 12 \] - If \( x = 1 \): \[ y = 2(1) - 6 = -4 \] **Solutions:** 1. \( x = 9 \) and \( y = 12 \) 2. \( x = 1 \) and \( y = -4 \) --- **4.2 State the arithmetic and geometric sequences for each of your solutions.** **First Solution (\( x = 9 \) and \( y = 12 \)):** - **Arithmetic Sequence (AP):** \[ 6, \, 9, \, 12 \] - **Common Difference (\( d \))**: \( 9 - 6 = 3 \) - **Geometric Sequence (GP):** \[ 9, \, 12, \, 16 \] - **Common Ratio (\( r \))**: \( \frac{12}{9} = \frac{4}{3} \) and \( \frac{16}{12} = \frac{4}{3} \) **Second Solution (\( x = 1 \) and \( y = -4 \)):** - **Arithmetic Sequence (AP):** \[ 6, \, 1, \, -4 \] - **Common Difference (\( d \))**: \( 1 - 6 = -5 \) and \( -4 - 1 = -5 \) - **Geometric Sequence (GP):** \[ 1, \, -4, \, 16 \] - **Common Ratio (\( r \))**: \( \frac{-4}{1} = -4 \) and \( \frac{16}{-4} = -4 \) **Summary of Solutions:** 1. **First Solution:** - **Sequence:** \( 6; 9; 12; 16 \) - **AP:** \( 6, 9, 12 \) with common difference \( d = 3 \) - **GP:** \( 9, 12, 16 \) with common ratio \( r = \frac{4}{3} \) 2. **Second Solution:** - **Sequence:** \( 6; 1; -4; 16 \) - **AP:** \( 6, 1, -4 \) with common difference \( d = -5 \) - **GP:** \( 1, -4, 16 \) with common ratio \( r = -4 \)

Reviewed and approved by the UpStudy tutoring team

error msg
Explain
Simplify this solution

Extra Insights

To find \( x \) and \( y \), we know that in an arithmetic progression, the difference between consecutive terms is consistent. Therefore, we can write the equation for \( x \) as: \[ x = 6 + d \] where \( d \) is the common difference. Then for \( y \): \[ y = x + d = (6 + d) + d = 6 + 2d \] Given that the last three terms \( x \), \( y \), and \( 16 \) form a geometric progression, the ratio between consecutive terms must be equal. Thus we have: \[ \frac{y}{x} = \frac{16}{y} \] Cross-multiplying gives: \[ y^2 = 16x \] Now substituting \( x \) and \( y \) in terms of \( d \): 1. From \( y = 6 + 2d \): \( (6 + 2d)^2 = 16(6 + d) \) Simplifying: \( 36 + 24d + 4d^2 = 96 + 16d \) Rearranging to form a quadratic equation: \( 4d^2 + 8d - 60 = 0 \) Dividing by 4: \( d^2 + 2d - 15 = 0 \) Factoring gives us: \( (d + 5)(d - 3) = 0 \) Thus, \( d = -5 \) or \( d = 3 \). 2. If \( d = 3 \): \( x = 6 + 3 = 9 \) \( y = 6 + 2(3) = 12 \) 3. If \( d = -5 \): \( x = 6 - 5 = 1 \) \( y = 6 + 2(-5) = -4 \) Thus, we have two sets of values for \( (x, y) \): \( (9, 12) \) and \( (1, -4) \). For the two cases: If \( (x, y) = (9, 12) \): - Arithmetic sequence: \( 6, 9, 12, 16 \) - Geometric sequence: \( 9, 12, 16 \) with common ratio \( \frac{12}{9} = \frac{16}{12} = \frac{4}{3} \). If \( (x, y) = (1, -4) \): - Arithmetic sequence: \( 6, 1, -4, 16 \) - Geometric sequence: \( 1, -4, 16 \) with common ratio \( \frac{-4}{1} = \frac{16}{-4} = -4 \).

Related Questions

5. \( \begin{aligned} j-9 \pi= & 12 \pi \\ +9 \pi & +9 \pi\end{aligned} \)
Algebra United States Jan 22, 2025
4. \( \frac{2 x-1}{3}-\frac{4 x-8}{6} \) 6. \( \frac{3}{x-1}+\frac{1}{1-x} \) 8. \( \frac{x}{x+y}+\frac{y}{x-y} \) 10. \( \frac{x+3}{x-3}-\frac{x-3}{x+3} \) 12. \( \frac{5}{3(a-b)}+\frac{3}{2(b-a)} \) 14. \( \frac{x}{x^{2}-y^{2}}+\frac{y}{y^{2}-x^{2}} \) 16. \( \frac{4 x}{(x+y)^{2}}-\frac{4}{x+y} \) 18. \( \frac{3 a}{9 a^{2}-4 b^{2}}-\frac{1}{3 a+2 b} \) 20. \( \frac{2 y}{(x-2 y)^{2}}+\frac{1}{x-2 y} \)
Algebra United States Jan 22, 2025
1. Find the products. \( \begin{array}{ll}\text { a) }(x+2)(x+3) & \text { (b) }(p-3)(p-4) \\ \text { c) }(a-2)(a+3) & \text { d) }(2 x-3)(3 x+1) \\ \text { e) }(2 y+5)(3 y-2) & \text { f) }(3 a-7)(2 a+1) \\ \text { g) }(2 a-3)(3 a-2) & \text { (h) }(4 y+1)(7 y-2) \\ \text { i) }(x-3)(x-2) & \text { j) }(4 a-2)(3 a+5) \\ \text { k) }(2 x-3 y)(2 x+4 y) & \text { 1) }(2 a-7 b)(3 a-2 b) \\ \text { m) }(7 x+3 y)(2 x-4 y) & \text { n) }(3 a-2 b)(2 a+11 b)\end{array} \)
Algebra South Africa Jan 22, 2025
\begin{tabular}{l} (a) Simplify \( \frac{2}{y^{0}} \) \\ (b) Simplify fully \( \left(16 a^{4}\right)^{\frac{3}{4}} \) \\ 5 \\ Factorise fully \( 18 c^{3} d^{2}-21 c^{2} \) \\ (Total for Question 4 is 3 marks) \\ \hline\end{tabular}
Algebra United Arab Emirates Jan 22, 2025
1. Find the products. a) \( (x+2)(x+3) \) c) \( (a-2)(a+3) \) e) \( (2 y+5)(3 y-2) \) g) \( (2 a-3)(3 a-2) \) i) \( (x-3)(x-2) \) k) \( (2 x-3 y)(2 x+4 y) \) m) \( (7 x+3 y)(2 x-4 y) \)
Algebra South Africa Jan 22, 2025

Latest Algebra Questions

2. Find the products. \( \begin{array}{ll}\text { a) }(x+1)(x+3) & \text { b) }(a+3)(a+4) \\ \text { c) }\left(x^{2}-1\right)\left(2 x^{2}+3\right) & \text { d) }\left(a^{2}-2\right)\left(a^{2}+2\right) \\ \text { e) }(3 x y+2 z)(2 x y-3 z) & \text { f) }(p r+2 t)(3 p r-t) \\ \text { g) }\left(x^{2}-4\right)\left(4 x^{2}-3\right) & \text { h) }\left(2 a^{2}+1\right)\left(3 a^{2}+1\right) \\ \text { i) }\left(x+\frac{1}{2}\right)\left(x+\frac{1}{4}\right) & \text { j) }\left(2 x+\frac{1}{3}\right)\left(3 x-\frac{1}{2}\right) \\ \text { k) }(2 a-b)(2 a+b) & \text { l) }(3 x+2 y)(3 x-2 y) \\ \text { m) }\left(\frac{1}{2} x+\frac{1}{3} y\right)\left(\frac{1}{3} x-\frac{1}{2} y\right) & \text {-n) }(x+3)^{2}\end{array} \)
Algebra South Africa Jan 22, 2025
5. \( \begin{aligned} j-9 \pi= & 12 \pi \\ +9 \pi & +9 \pi\end{aligned} \)
Algebra United States Jan 22, 2025
7 Expand and simplify \( 5 x(3 x+4)(2 x-1) \)
Algebra United Arab Emirates Jan 22, 2025
1. Find the products. \( \begin{array}{ll}\text { a) }(x+2)(x+3) & \text { (b) }(p-3)(p-4) \\ \text { c) }(a-2)(a+3) & \text { d) }(2 x-3)(3 x+1) \\ \text { e) }(2 y+5)(3 y-2) & \text { f) }(3 a-7)(2 a+1) \\ \text { g) }(2 a-3)(3 a-2) & \text { (h) }(4 y+1)(7 y-2) \\ \text { i) }(x-3)(x-2) & \text { j) }(4 a-2)(3 a+5) \\ \text { k) }(2 x-3 y)(2 x+4 y) & \text { 1) }(2 a-7 b)(3 a-2 b) \\ \text { m) }(7 x+3 y)(2 x-4 y) & \text { n) }(3 a-2 b)(2 a+11 b)\end{array} \)
Algebra South Africa Jan 22, 2025
1. Find the products. \( \begin{array}{ll}\text { a) }(x+2)(x+3) & \text { \b) }(p-3)(p-4) \\ \text { c) }(a-2)(a+3) & \text { d) }(2 x-3)(3 x+1) \\ \text { e) }(2 y+5)(3 y-2) & \text { f) }(3 a-7)(2 a+1) \\ \text { g) }(2 a-3)(3 a-2) & \text { (h) }(4 y+1)(7 y-2) \\ \text { i) }(x-3)(x-2) & \text { j) }(4 a-2)(3 a+5) \\ \text { k) }(2 x-3 y)(2 x+4 y) & \text { \l) }(2 a-7 b)(3 a-2 b) \\ \text { m) }(7 x+3 y)(2 x-4 y) & \text { n) }(3 a-2 b)(2 a+11 b) \\ \text { 2. Find the products. } & \text { b) }(a+3)(a+4) \\ \text { a) }(x+1)(x+3) & \text { (d) }\left(a^{2}-2\right)\left(a^{2}+2\right) \\ \text { c) }\left(x^{2}-1\right)\left(2 x^{2}+3\right) & \text { f) }(p r+2 t)(3 p r-t) \\ \text { (e) }(3 x y+2 z)(2 x y-3 z) & \text { h) }\left(2 a^{2}+1\right)\left(3 a^{2}+1\right) \\ \text { g) }\left(x^{2}-4\right)\left(4 x^{2}-3\right) & \text { j) }\left(2 x+\frac{1}{3}\right)\left(3 x-\frac{1}{2}\right) \\ \text { i) }\left(x+\frac{1}{2}\right)\left(x+\frac{1}{4}\right) & \text { l) }(3 x+2 y)(3 x-2 y) \\ \text { k) }(2 a-b)(2 a+b) & \text { n) }(x+3)^{2} \\ \text { m) }\left(\frac{1}{2} x+\frac{1}{3} y\right)\left(\frac{1}{3} x-\frac{1}{2} y\right) & \end{array} \)
Algebra South Africa Jan 22, 2025
Prev Next
Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy