Math Problem Statement

It is given that 8; 2π‘₯; 2𝑦 forms an arithmetic sequence and 2π‘₯; 2𝑦; 36 forms a geometric sequence. 2.1.1 By using the arithmetic sequence, show that 𝑦 = 2π‘₯ βˆ’ 4. (2) 2.1.2 Hence, determine the values of π‘₯ and 𝑦.

Solution

Let's break down the problem step by step.

Problem Recap:

  • We have an arithmetic sequence: 8,2x,2y8, 2x, 2y
  • We have a geometric sequence: 2x,2y,362x, 2y, 36

2.1.1 Show that y=2xβˆ’4y = 2x - 4 using the arithmetic sequence.

Step 1: Understand the condition of an arithmetic sequence.

In an arithmetic sequence, the difference between consecutive terms is constant. So, we can write: 2xβˆ’8=2yβˆ’2x2x - 8 = 2y - 2x

Step 2: Simplify the equation.

Let's solve the equation: 2xβˆ’8=2yβˆ’2x2x - 8 = 2y - 2x Add 2x2x to both sides: 4xβˆ’8=2y4x - 8 = 2y Divide the whole equation by 2: 2xβˆ’4=y2x - 4 = y Thus, we have shown that: y=2xβˆ’4y = 2x - 4

2.1.2 Determine the values of xx and yy.

Step 1: Use the geometric sequence condition.

In a geometric sequence, the ratio between consecutive terms is constant. Therefore: 2y2x=362y\frac{2y}{2x} = \frac{36}{2y} Simplifying this: yx=18y\frac{y}{x} = \frac{18}{y} Cross-multiply to get: y2=18xy^2 = 18x

Step 2: Substitute the expression for yy from 2.1.1.

We know that y=2xβˆ’4y = 2x - 4. Substitute this into the equation: (2xβˆ’4)2=18x(2x - 4)^2 = 18x

Step 3: Expand and simplify the quadratic equation.

Expanding: 4x2βˆ’16x+16=18x4x^2 - 16x + 16 = 18x Rearrange to form a standard quadratic equation: 4x2βˆ’34x+16=04x^2 - 34x + 16 = 0 Divide the entire equation by 2: 2x2βˆ’17x+8=02x^2 - 17x + 8 = 0

Step 4: Solve the quadratic equation.

Use the quadratic formula: x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} Here, a=2a = 2, b=βˆ’17b = -17, and c=8c = 8. Substituting these values: x=17Β±(βˆ’17)2βˆ’4(2)(8)4x = \frac{17 \pm \sqrt{(-17)^2 - 4(2)(8)}}{4} x=17Β±289βˆ’644x = \frac{17 \pm \sqrt{289 - 64}}{4} x=17Β±2254x = \frac{17 \pm \sqrt{225}}{4} x=17Β±154x = \frac{17 \pm 15}{4}

So, the two possible values for xx are: x=324=8orx=24=0.5x = \frac{32}{4} = 8 \quad \text{or} \quad x = \frac{2}{4} = 0.5

Step 5: Find the corresponding values of yy.

If x=8x = 8: y=2(8)βˆ’4=16βˆ’4=12y = 2(8) - 4 = 16 - 4 = 12

If x=0.5x = 0.5: y=2(0.5)βˆ’4=1βˆ’4=βˆ’3y = 2(0.5) - 4 = 1 - 4 = -3

Thus, the possible values for xx and yy are:

  • x=8x = 8, y=12y = 12
  • x=0.5x = 0.5, y=βˆ’3y = -3

Conclusion:

The values of xx and yy can be (8,12)(8, 12) or (0.5,βˆ’3)(0.5, -3).

Would you like more details or have any questions about this solution?


  1. What are the properties of arithmetic sequences?
  2. How do you derive the general term of an arithmetic sequence?
  3. What defines a geometric sequence, and how do you find its common ratio?
  4. How can the quadratic formula be used to solve different types of quadratic equations?
  5. In what other contexts might these types of sequences appear?

Tip: When working with sequences, always check if there's more than one possible solution, as seen here with both positive and negative values for xx and yy.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Sequences
Geometric Sequences
Quadratic Equations

Formulas

Quadratic formula

Theorems

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Suitable Grade Level

Grades 9-12