Understanding the Problem: AP, GP, and Logarithms

The problem asks us to determine the relationship between three numbers, x, y, and z, given three conditions involving geometric progression (GP), arithmetic progression (AP), and logarithms. We are given:

1. \(\log_x y, \log_z x, \log_y z\) are in Geometric Progression (GP).
2. \(xyz = 64\).
3. \(x^3, y^3, z^3\) are in Arithmetic Progression (AP).

Let's analyze each condition to find the values or relationships between x, y, and z.

Analyzing Condition 1: Logarithmic Terms in GP

If three terms \(a, b, c\) are in GP, then the square of the middle term equals the product of the other two terms, i.e., \(b^2 = ac\).

Applying this to the given logarithmic terms:

$$(\log_z x)^2 = (\log_x y) \times (\log_y z)$$

We can use the change of base formula for logarithms, which states that \(\log_b a = \frac{\log_c a}{\log_c b}\) for any valid base \(c\). Let's use the natural logarithm (base \(e\)) or any common base.

$$ \left(\frac{\ln x}{\ln z}\right)^2 = \left(\frac{\ln y}{\ln x}\right) \times \left(\frac{\ln z}{\ln y}\right) $$

Assuming \(\ln y \neq 0\) (which means \(y \neq 1\)), we can cancel \(\ln y\) from the right side:

$$ \frac{(\ln x)^2}{(\ln z)^2} = \frac{\ln z}{\ln x} $$

Now, cross-multiply:

$$ (\ln x)^2 \times (\ln x) = (\ln z) \times (\ln z)^2 $$

$$ (\ln x)^3 = (\ln z)^3 $$

Taking the cube root of both sides:

$$ \ln x = \ln z $$

Exponentiating both sides with base \(e\):

$$ e^{\ln x} = e^{\ln z} $$

$$ x = z $$

So, from the first condition, we find that x must be equal to z.

Analyzing Condition 3: Cubed Terms in AP

If three terms \(a, b, c\) are in AP, then twice the middle term equals the sum of the other two terms, i.e., \(2b = a + c\).

Applying this to the given terms \(x^3, y^3, z^3\):

$$ 2y^3 = x^3 + z^3 $$

From our analysis of Condition 1, we found that \(x = z\). Substitute \(z\) with \(x\) in this equation:

$$ 2y^3 = x^3 + x^3 $$

$$ 2y^3 = 2x^3 $$

Divide both sides by 2:

$$ y^3 = x^3 $$

Taking the cube root of both sides:

$$ y = x $$

So, from the third condition and the result of the first, we find that y must be equal to x.

Combining Results and Using Condition 2

From Condition 1, we got \(x = z\). From Condition 3 (using \(x=z\)), we got \(y = x\).

Combining these results, we have \(x = y = z\).

Now, let's use Condition 2: \(xyz = 64\).

Substitute \(x\) for \(y\) and \(z\):

$$ x \times x \times x = 64 $$

$$ x^3 = 64 $$

To find x, take the cube root of 64:

$$ x = \sqrt[3]{64} $$

$$ x = 4 $$

Since \(x = y = z\), we have \(x = y = z = 4\).

Checking if x, y, and z are in AP and GP

We found that \(x = 4\), \(y = 4\), and \(z = 4\). Let's check if these numbers are in AP and GP.

Check for AP:
For numbers to be in AP, the difference between consecutive terms must be constant. Is \(y - x = z - y\)?

$$ 4 - 4 = 4 - 4 $$

$$ 0 = 0 $$

Yes, x, y, and z (4, 4, 4) are in AP with a common difference of 0.

Check for GP:
For numbers to be in GP, the ratio between consecutive terms must be constant. Is \(y/x = z/y\)? (Assuming x, y, z are non-zero, which they are, as 4).

$$ \frac{4}{4} = \frac{4}{4} $$

$$ 1 = 1 $$

Yes, x, y, and z (4, 4, 4) are in GP with a common ratio of 1.

Conclusion

Since x, y, and z are 4, 4, and 4, they satisfy the conditions for both Arithmetic Progression and Geometric Progression.

Revision Table: Key Concepts

Additional Information: Properties of Sequences

Sequences of numbers can follow various patterns. AP and GP are two fundamental types.

• Arithmetic Progression: Terms are formed by adding a constant value (the common difference) to the previous term. Example: 2, 5, 8, 11... (common difference is 3).
• Geometric Progression: Terms are formed by multiplying the previous term by a constant value (the common ratio). Example: 3, 6, 12, 24... (common ratio is 2).

An interesting case is when all terms in a sequence are the same (e.g., 4, 4, 4). Let's see why this fits both definitions:

• Is 4, 4, 4 in AP? The difference between consecutive terms is \(4 - 4 = 0\). Since the difference is constant (0), it is an AP with a common difference of 0.
• Is 4, 4, 4 in GP? The ratio between consecutive terms is \(4/4 = 1\). Since the ratio is constant (1), it is a GP with a common ratio of 1 (provided the terms are non-zero).

Therefore, a sequence of identical, non-zero numbers is always both an AP and a GP. This aligns with our finding that \(x=y=z=4\).