Analyzing Progression Types Based on Geometric Progression

The question asks us to determine which of the given statements is/are correct, given that three numbers \(x\), \(y\), and \(z\) are in Geometric Progression (GP).

If \(x\), \(y\), \(z\) are in GP, this means that the ratio of consecutive terms is constant. Mathematically, this is expressed as:

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

Cross-multiplying, we get the fundamental property of three terms in GP:

\(y^2 = xz\)

We will use this property to evaluate each statement.

Statement 1: Are \(\ln(3x)\), \(\ln(3y)\), \(\ln(3z)\) in AP?

For three terms \(a\), \(b\), \(c\) to be in Arithmetic Progression (AP), the middle term must be the average of the other two. This can be written as \(2b = a + c\).

In this case, the terms are \(a = \ln(3x)\), \(b = \ln(3y)\), and \(c = \ln(3z)\). For them to be in AP, the following condition must hold:

\(2 \ln(3y) = \ln(3x) + \ln(3z)\)

Using the properties of logarithms, \(\ln(A) + \ln(B) = \ln(AB)\) and \(k \ln(A) = \ln(A^k)\), we can rewrite the equation:

Left side: \(2 \ln(3y) = \ln((3y)^2) = \ln(9y^2)\)

Right side: \(\ln(3x) + \ln(3z) = \ln(3x \cdot 3z) = \ln(9xz)\)

So, the condition for the terms to be in AP becomes:

\(\ln(9y^2) = \ln(9xz)\)

Since the natural logarithm function is one-to-one, this equality holds if and only if their arguments are equal:

\(9y^2 = 9xz\)

Dividing both sides by 9 (assuming \(x, y, z\) are positive for \(\ln\) to be defined), we get:

\(y^2 = xz\)

This is exactly the condition we derived from the fact that \(x\), \(y\), \(z\) are in GP. Since \(y^2 = xz\) is true, the condition \(2 \ln(3y) = \ln(3x) + \ln(3z)\) is also true.

Therefore, statement 1 is correct.

Statement 2: Are \(xyz + \ln(x)\), \(xyz + \ln(y)\), \(xyz + \ln(z)\) in HP?

For three terms \(a\), \(b\), \(c\) to be in Harmonic Progression (HP), their reciprocals (\(1/a\), \(1/b\), \(1/c\)) must be in Arithmetic Progression (AP).

In this case, the terms are \(a = xyz + \ln(x)\), \(b = xyz + \ln(y)\), and \(c = xyz + \ln(z)\). For them to be in HP, their reciprocals must be in AP:

\(\frac{1}{xyz + \ln(x)}\), \(\frac{1}{xyz + \ln(y)}\), \(\frac{1}{xyz + \ln(z)}\) must be in AP.

Applying the AP condition \(2b = a + c\) to the reciprocals:

\(2 \left( \frac{1}{xyz + \ln(y)} \right) = \frac{1}{xyz + \ln(x)} + \frac{1}{xyz + \ln(z)}\)

Simplify the right side by finding a common denominator:

\(\frac{2}{xyz + \ln(y)} = \frac{(xyz + \ln(z)) + (xyz + \ln(x))}{(xyz + \ln(x))(xyz + \ln(z))}\)

\(\frac{2}{xyz + \ln(y)} = \frac{2xyz + \ln(x) + \ln(z)}{(xyz + \ln(x))(xyz + \ln(z))}\)

We know that if \(x\), \(y\), \(z\) are in GP, then \(y^2 = xz\). Taking the natural logarithm of both sides (assuming positive \(x, y, z\)):

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

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

This shows that \(\ln(x)\), \(\ln(y)\), \(\ln(z)\) are in AP.

Now substitute \(\ln(x) + \ln(z) = 2\ln(y)\) into the HP condition equation:

\(\frac{2}{xyz + \ln(y)} = \frac{2xyz + 2\ln(y)}{(xyz + \ln(x))(xyz + \ln(z))}\)

\(\frac{2}{xyz + \ln(y)} = \frac{2(xyz + \ln(y))}{(xyz + \ln(x))(xyz + \ln(z))}\)

Assuming \(xyz + \ln(y) \neq 0\) and \(xyz + \ln(x)\), \(xyz + \ln(z)\) are defined and non-zero, we can divide both sides by 2 and cross-multiply:

\(1 \cdot (xyz + \ln(x))(xyz + \ln(z)) = (xyz + \ln(y)) \cdot (xyz + \ln(y))\)

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

Expand both sides:

\((xyz)^2 + xyz \ln(z) + xyz \ln(x) + \ln(x)\ln(z) = (xyz)^2 + 2 \cdot xyz \ln(y) + (\ln(y))^2\)

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

Substitute \(\ln(x) + \ln(z) = 2\ln(y)\) into the equation:

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

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

Subtracting \((xyz)^2 + 2xyz \ln(y)\) from both sides, we are left with the condition:

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

This condition means that \(\ln(x)\), \(\ln(y)\), \(\ln(z)\) are in GP.

However, we know that if \(x\), \(y\), \(z\) are in GP, then \(\ln(x)\), \(\ln(y)\), \(\ln(z)\) are in AP (\(2\ln(y) = \ln(x) + \ln(z)\)), not necessarily in GP (\(\ln(y)^2 = \ln(x)\ln(z)\)). For a sequence to be in both AP and GP, all its terms must be equal. Since \(\ln(x)\), \(\ln(y)\), \(\ln(z)\) are not necessarily equal (unless \(x=y=z\)), the condition \(\ln(x)\ln(z) = (\ln(y))^2\) is not generally true just because \(x\), \(y\), \(z\) are in GP.

Therefore, statement 2 is incorrect.

Conclusion

Based on our analysis:

• Statement 1 is correct: If \(x\), \(y\), \(z\) are in GP, then \(\ln(3x)\), \(\ln(3y)\), \(\ln(3z)\) are in AP.
• Statement 2 is incorrect: If \(x\), \(y\), \(z\) are in GP, then \(xyz + \ln(x)\), \(xyz + \ln(y)\), \(xyz + \ln(z)\) are not necessarily in HP.

The correct option is the one stating that only Statement 1 is correct.