Understanding Arithmetic Progression (AP)

An Arithmetic Progression (AP) is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference.

If three terms, say a, b, and c, are in AP, then the difference between the second and first term is equal to the difference between the third and second term. Mathematically, this can be written as:

\(\text{b} - \text{a} = \text{c} - \text{b}\)

Alternatively, rearranging the terms, we get \(2\text{b} = \text{a} + \text{c}\), which means the middle term is the average of the first and third terms.

Solving for x in the Given AP

The question states that the terms \(\text{x}^2\), \(\text{x}\), and \(-8\) are in AP. We can use the property of the common difference being constant to find the value(s) of x.

Let the terms be:

• First term (\(\text{a}_1\)): \(\text{x}^2\)
• Second term (\(\text{a}_2\)): \(\text{x}\)
• Third term (\(\text{a}_3\)): \(-8\)

According to the property of AP, the common difference between \(\text{a}_2\) and \(\text{a}_1\) must be equal to the common difference between \(\text{a}_3\) and \(\text{a}_2\).

\(\text{a}_2 - \text{a}_1 = \text{a}_3 - \text{a}_2\)

Substitute the given terms into this equation:

\(\text{x} - \text{x}^2 = -8 - \text{x}\)

Solving the Quadratic Equation

We need to solve the equation \(\text{x} - \text{x}^2 = -8 - \text{x}\) for x. Let's rearrange the terms to form a standard quadratic equation (\(\text{ax}^2 + \text{bx} + \text{c} = 0\)).

Move all terms to one side:

\(\text{x} - \text{x}^2 + 8 + \text{x} = 0\)

Combine like terms:

\(-\text{x}^2 + 2\text{x} + 8 = 0\)

To make the leading coefficient positive, multiply the entire equation by \(-1\):

\(\text{x}^2 - 2\text{x} - 8 = 0\)

Now we have a quadratic equation. We can solve this by factoring, completing the square, or using the quadratic formula. Factoring is often the quickest method if possible.

We look for two numbers that multiply to \(-8\) and add up to \(-2\). These numbers are \(4\) and \(-2\), but they add to \(2\). Let's try \(-4\) and \(2\). These multiply to \(-8\) and add up to \(-2\).

So, we can factor the quadratic equation as:

\((\text{x} - 4)(\text{x} + 2) = 0\)

For this equation to be true, one or both of the factors must be equal to zero.

• Case 1: \(\text{x} - 4 = 0\)
• \(\text{x} = 4\)

• Case 2: \(\text{x} + 2 = 0\)
• \(\text{x} = -2\)

So, the possible values for x are \(4\) and \(-2\).

Verifying the Solutions for AP

Let's check if these values of x actually result in the terms being in AP.

• If \(\text{x} = 4\): The terms are \(\text{4}^2\), \(4\), \(-8\), which simplifies to \(16\), \(4\), \(-8\).
  • Common difference 1: \(4 - 16 = -12\)
  • Common difference 2: \(-8 - 4 = -12\)
  Since the common differences are equal (\(-12\)), the terms \(16, 4, -8\) are in AP. So, \(\text{x} = 4\) is a valid solution.
• If \(\text{x} = -2\): The terms are \((-\text{2})^2\), \(-2\), \(-8\), which simplifies to \(4\), \(-2\), \(-8\).
  • Common difference 1: \(-2 - 4 = -6\)
  • Common difference 2: \(-8 - (-2) = -8 + 2 = -6\)
  Since the common differences are equal (\(-6\)), the terms \(4, -2, -8\) are in AP. So, \(\text{x} = -2\) is a valid solution.

Both values \(\text{x} = -2\) and \(\text{x} = 4\) satisfy the condition for the given terms to be in AP.

The set of possible values for x is \(\{-2, 4\}\).

Revision Table: Key AP Concepts

Additional Information: Solving Quadratic Equations

When you encounter a quadratic equation like \(\text{ax}^2 + \text{bx} + \text{c} = 0\), there are several methods to find the values of x (the roots).

• Factoring: As used in this solution, find two numbers that multiply to \(ac\) and add up to \(b\). Use these numbers to split the middle term (\(bx\)) and then factor by grouping.
• Quadratic Formula: This formula always works and is given by:

  \(\text{x} = \frac{-\text{b} \pm \sqrt{\text{b}^2 - 4\text{ac}}}{2\text{a}}\)

  For the equation \(\text{x}^2 - 2\text{x} - 8 = 0\), we have \(\text{a}=1\), \(\text{b}=-2\), \(\text{c}=-8\). Plugging these into the formula gives:

  \(\text{x} = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)}\)

  \(\text{x} = \frac{2 \pm \sqrt{4 + 32}}{2}\)

  \(\text{x} = \frac{2 \pm \sqrt{36}}{2}\)

  \(\text{x} = \frac{2 \pm 6}{2}\)

  This gives two solutions:

  \(\text{x}_1 = \frac{2 + 6}{2} = \frac{8}{2} = 4\)

  \(\text{x}_2 = \frac{2 - 6}{2} = \frac{-4}{2} = -2\)

  These are the same solutions obtained by factoring, confirming our result that \(\text{x} \in \{-2, 4\}\).
• Completing the Square: This method involves manipulating the equation to form a perfect square trinomial on one side.

Understanding these methods helps solve various problems involving quadratic equations that might arise in topics like Arithmetic Progression.