GP Condition Analysis

The problem states that three numbers, \(x\), \(y\), and \(z\), are greater than 1 (\(x > 1, y > 1, z > 1\)) and form a Geometric Progression (GP).

By the definition of a Geometric Progression, the ratio between consecutive terms is constant. This means:

\( \frac{y}{x} = \frac{z}{y} \)

Cross-multiplying gives us the fundamental property:

\( y^2 = xz \)

Logarithm Properties and AP Relationship

We are asked to analyze the sequence involving the natural logarithms (ln) of these terms: \(\ln x\), \(\ln y\), and \(\ln z\). Since \(x, y, z\) are all greater than 1, their natural logarithms are defined and are real numbers.

Let's apply the natural logarithm to both sides of the GP property \(y^2 = xz\):

\( \ln(y^2) = \ln(xz) \)

Using the standard logarithm properties, specifically \(\ln(a^b) = b \ln a\) and \(\ln(ab) = \ln a + \ln b\), we can rewrite the equation:

\( 2 \ln y = \ln x + \ln z \)

Rearranging this equation to see the difference between consecutive terms:

\( \ln y - \ln x = \ln z - \ln y \)

This equality confirms that the difference between consecutive logarithmic terms (\(\ln y - \ln x\) and \(\ln z - \ln y\)) is constant. Therefore, the terms \(\ln x\), \(\ln y\), \(\ln z\) form an Arithmetic Progression (AP).

Understanding Progressions (AP, GP, HP)

To clarify the nature of the sequence we are analyzing, let's recall the definitions:

• Arithmetic Progression (AP): A sequence where the difference between any two successive members is constant. Example: 3, 5, 7, 9... (common difference is 2).
• Geometric Progression (GP): A sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. Example: 2, 6, 18, 54... (common ratio is 3).
• Harmonic Progression (HP): A sequence whose reciprocals form an Arithmetic Progression. Example: 1/5, 1/8, 1/11... (reciprocals 5, 8, 11 form an AP).

Analyzing the Target Sequence

The sequence we need to classify is:

\( A = \frac{1}{1+\ln x}, \quad B = \frac{1}{1+\ln y}, \quad C = \frac{1}{1+\ln z} \)

According to the definition of Harmonic Progression, we should check if the reciprocals of these terms form an Arithmetic Progression.

Let's find the reciprocals:

\( \frac{1}{A} = 1+\ln x \)

\( \frac{1}{B} = 1+\ln y \)

\( \frac{1}{C} = 1+\ln z \)

Verifying the Reciprocal Sequence is an AP

We already established that \(\ln x\), \(\ln y\), \(\ln z\) form an AP. Now, let's consider the sequence formed by adding 1 to each term: \(1+\ln x\), \(1+\ln y\), \(1+\ln z\).

Let's calculate the difference between consecutive terms in this reciprocal sequence:

Difference 1: \( \left(1+\ln y\right) - \left(1+\ln x\right) = 1 + \ln y - 1 - \ln x = \ln y - \ln x \)

Difference 2: \( \left(1+\ln z\right) - \left(1+\ln y\right) = 1 + \ln z - 1 - \ln y = \ln z - \ln y \)

Since we proved that \(\ln y - \ln x = \ln z - \ln y\) (because \(\ln x\), \(\ln y\), \(\ln z\) are in AP), the differences between the consecutive terms of the reciprocal sequence are equal.

This confirms that the sequence \(1+\ln x\), \(1+\ln y\), \(1+\ln z\) is indeed an Arithmetic Progression.

Conclusion on the Sequence Type

Given that the reciprocals of the terms \(\frac{1}{1+\ln x}\), \(\frac{1}{1+\ln y}\), \(\frac{1}{1+\ln z}\) form an Arithmetic Progression, the original sequence must be a Harmonic Progression (HP).