Understanding Progressions: AP, GP, and HP

This question involves three important types of sequences or progressions: Arithmetic Progression (AP), Geometric Progression (GP), and Harmonic Progression (HP). Let's first recall the defining property of each progression for three numbers, say a, b, and c.

• Arithmetic Progression (AP): Three numbers a, b, c are in AP if the middle term is the arithmetic mean of the other two. Mathematically, this means $2b = a + c$. The difference between consecutive terms is constant ($b-a = c-b$).
• Geometric Progression (GP): Three numbers a, b, c are in GP if the middle term is the geometric mean of the other two. Mathematically, this means \(b^2 = ac\). The ratio of consecutive terms is constant ($b/a = c/b$).
• Harmonic Progression (HP): Three numbers a, b, c are in HP if their reciprocals $1/a, 1/b, 1/c$ are in AP. Mathematically, this means \(2 \times \frac{1}{b} = \frac{1}{a} + \frac{1}{c}\), which simplifies to \(\frac{2}{b} = \frac{a+c}{ac}\), or \(b = \frac{2ac}{a+c}\). The middle term is the harmonic mean of the other two.

Analyzing the Given Information

We are given two pieces of information about the numbers x and z:

1. x, 3/2, z are in AP.
2. x, 3, z are in GP.

Applying the AP Property

Since x, 3/2, z are in AP, using the property $2b = a+c$, we have:

$$2 \times \frac{3}{2} = x + z$$

$$3 = x + z \quad \text{(Equation 1)}$$

Applying the GP Property

Since x, 3, z are in GP, using the property \(b^2 = ac\), we have:

$$3^2 = xz$$

$$9 = xz \quad \text{(Equation 2)}$$

Finding the Relationship for HP

We need to find which of the given options is in HP. Let's consider the general form of an option: x, k, z, where k is the middle term from the options (6, 4, 2, or 1).

If x, k, z are in HP, then their reciprocals $1/x, 1/k, 1/z$ must be in AP. Applying the AP property to the reciprocals:

$$2 \times \frac{1}{k} = \frac{1}{x} + \frac{1}{z}$$

$$\frac{2}{k} = \frac{z + x}{xz}$$

Now, we can use the results from Equation 1 ($x+z=3$) and Equation 2 ($xz=9$) to find the value that k must take if x, k, z are in HP.

Substitute $x+z=3$ and $xz=9$ into the equation for k:

$$\frac{2}{k} = \frac{3}{9}$$

$$\frac{2}{k} = \frac{1}{3}$$

Now, solve for k:

$$2 \times 3 = k \times 1$$

$$6 = k$$

Conclusion: Identifying the HP Sequence

Our calculation shows that for x, k, z to be in HP, the middle term k must be 6. We now look at the options provided to see which one has 6 as the middle term.

• Option 1: x, 6, z
• Option 2: x, 4, z
• Option 3: x, 2, z
• Option 4: x, 1, z

Comparing our result $k=6$ with the options, we find that Option 1 matches the condition for x, k, z to be in HP based on the initial conditions for x and z.

Therefore, x, 6, z will be in HP.

Revision Table: Key Progression Concepts

Additional Information on AP, GP, and HP Relationships

There are interesting relationships between the means:

• For any two positive numbers a and c, their Arithmetic Mean (AM), Geometric Mean (GM), and Harmonic Mean (HM) satisfy the inequality \(AM \ge GM \ge HM\).
• If a, b, c are in GP, then a, b, c are the Geometric Means between certain terms in other sequences.
• If three numbers are in GP, the sequence formed by their reciprocals is also in GP. However, if three numbers are in AP, their reciprocals are in HP, and vice-versa.
• In this problem, we found that $x+z=3$ and $xz=9$. The quadratic equation \(t^2 - (x+z)t + xz = 0\) becomes \(t^2 - 3t + 9 = 0\). The roots are \(t = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(9)}}{2} = \frac{3 \pm \sqrt{9 - 36}}{2} = \frac{3 \pm \sqrt{-27}}{2} = \frac{3 \pm 3i\sqrt{3}}{2}\). So, x and z are complex numbers in this specific case, but the relationships derived ($x+z=3$ and $xz=9$) hold true and are used to find the HP middle term. The formulas for AP, GP, and HP apply to complex numbers as well.