If `x,(3)/(2),z" are in AP, "x,3,z` are in GP, then which one of the following will be in HP?

Options

A. x,6,z
B. x,4,z
C. x,2,z
D. x,1,z

Explanation

To solve the problem, we need to analyze the conditions given for the sequences and determine which set of numbers will be in Harmonic Progression (HP).

### Given:
1. \( x, \frac{3}{2}, z \) are in Arithmetic Progression (AP).
2. \( x, 3, z \) are in Geometric Progression (GP).

### Step 1: Use the AP condition
For three numbers \( a, b, c \) to be in AP, the condition is:
\[ 2b = a + c \]

Here, substituting \( a = x \), \( b = \frac{3}{2} \), and \( c = z \):
\[ 2 \cdot \frac{3}{2} = x + z \]
\[ 3 = x + z \]
Thus, we have:
\[ x + z = 3 \quad \text{(Equation 1)} \]

### Step 2: Use the GP condition
For three numbers \( a, b, c \) to be in GP, the condition is:
\[ b^2 = ac \]

Here, substituting \( a = x \), \( b = 3 \), and \( c = z \):
\[ 3^2 = x \cdot z \]
\[ 9 = xz \quad \text{(Equation 2)} \]

### Step 3: Solve the equations
We have two equations:
1. \( x + z = 3 \) (Equation 1)
2. \( xz = 9 \) (Equation 2)

From Equation 1, we can express \( z \) in terms of \( x \):
\[ z = 3 - x \]

Now substitute \( z \) into Equation 2:
\[ x(3 - x) = 9 \]
Expanding this gives:
\[ 3x - x^2 = 9 \]
Rearranging it leads to:
\[ x^2 - 3x + 9 = 0 \]

### Step 4: Find the roots of the quadratic equation
Now we can find the roots using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -3, c = 9 \):
\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} \]
\[ x = \frac{3 \pm \sqrt{9 - 36}}{2} \]
\[ x = \frac{3 \pm \sqrt{-27}}{2} \]
\[ x = \frac{3 \pm 3i\sqrt{3}}{2} \]

### Step 5: Find \( z \)
Using \( z = 3 - x \):
\[ z = 3 - \frac{3 \pm 3i\sqrt{3}}{2} \]
This gives us complex values for \( x \) and \( z \).

### Step 6: Check which set is in HP
To check if \( x, 6, z \) are in HP, we use the condition:
\[ \frac{2}{6} = \frac{1}{x} + \frac{1}{z} \]

Calculating:
\[ \frac{1}{3} = \frac{1}{x} + \frac{1}{z} \]
Using \( x + z = 3 \) and \( xz = 9 \):
\[ \frac{1}{x} + \frac{1}{z} = \frac{z + x}{xz} = \frac{3}{9} = \frac{1}{3} \]

Since both sides are equal, \( x, 6, z \) are indeed in HP.

### Conclusion
Thus, the numbers \( x, 6, z \) are in Harmonic Progression (HP).

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