Understanding the Problem: Exponents, GP, and AP

We are given an equation relating three variables $p$, $q$, and $r$ through exponents: $p^x = q^y = r^z$. We are also told that the exponents $x$, $y$, and $z$ are in a Geometric Progression (GP). Our task is to determine which of the two given statements about $p$, $q$, $r$ and their logarithms is correct.

Analyzing the Given Information

Let's represent the common relationship $p^x = q^y = r^z$ by a constant $k$. So, we have:

• $p^x = k \implies p = k^{1/x}$
• $q^y = k \implies q = k^{1/y}$
• $r^z = k \implies r = k^{1/z}$

We are also given that $x$, $y$, and $z$ are in Geometric Progression (GP). This means the ratio between consecutive terms is constant, i.e., $\frac{y}{x} = \frac{z}{y}$. Cross-multiplying gives us the condition for $x, y, z$ being in GP: $y^2 = xz$.

Evaluating Statement I: $p$, $q$, $r$ are in AP

For $p$, $q$, and $r$ to be in Arithmetic Progression (AP), the difference between consecutive terms must be constant. This means $q - p = r - q$, or $2q = p + r$. Let's substitute the expressions for $p$, $q$, and $r$ in terms of $k$:

We have $p = k^{1/x}$, $q = k^{1/y}$, $r = k^{1/z}$. From $y^2 = xz$, we can take the reciprocal of both sides: $\frac{1}{y^2} = \frac{1}{xz}$. This can be rewritten as $\left(\frac{1}{y}\right)^2 = \frac{1}{x} \cdot \frac{1}{z}$. This implies that the terms $\frac{1}{x}$, $\frac{1}{y}$, and $\frac{1}{z}$ are in Geometric Progression (GP).

Now let's consider the terms $p$, $q$, $r$. Using the property $q = k^{1/y}$, we can write:

$q^2 = (k^{1/y})^2 = k^{2/y}$

Since $\frac{1}{y}$ is the geometric mean of $\frac{1}{x}$ and $\frac{1}{z}$, we have $\frac{2}{y} = \frac{1}{x} + \frac{1}{z}$ is NOT necessarily true. However, we found that $(\frac{1}{y})^2 = \frac{1}{x} \cdot \frac{1}{z}$. So, $q^2 = k^{2/y}$. Let's relate this to $p$ and $r$. $p \cdot r = k^{1/x} \cdot k^{1/z} = k^{(1/x + 1/z)}$. We know $y^2 = xz$. Let's rewrite the exponents: $p = k^{1/x}$ $q = k^{1/y}$ $r = k^{1/z}$ Since $1/x, 1/y, 1/z$ are in GP, let $1/y = \lambda (1/x)$ and $1/z = \lambda (1/y) = \lambda^2 (1/x)$ for some common ratio $\lambda$. Then $p = k^{1/x}$ $q = k^{\lambda/x} = (k^{1/x})^\lambda = p^\lambda$ $r = k^{\lambda^2/x} = (k^{1/x})^{\lambda^2} = p^{\lambda^2}$ This shows that $p, q, r$ themselves form a GP ($q/p = r/q$), not an AP ($q-p = r-q$). Therefore, Statement I is incorrect.

Evaluating Statement II: $\ln p$, $\ln q$, $\ln r$ are in GP

Let's find the natural logarithms ($\ln$) of $p$, $q$, and $r$. We assume $k > 0$ and $k \neq 1$ for the logarithms to be well-defined and non-zero.

• $\ln p = \ln(k^{1/x}) = \frac{1}{x} \ln k$
• $\ln q = \ln(k^{1/y}) = \frac{1}{y} \ln k$
• $\ln r = \ln(k^{1/z}) = \frac{1}{z} \ln k$

For these three logarithmic terms to be in Geometric Progression (GP), the square of the middle term must equal the product of the other two terms:

$(\ln q)^2 = (\ln p)(\ln r)$

Substitute the expressions:

$\left(\frac{1}{y} \ln k\right)^2 = \left(\frac{1}{x} \ln k\right) \left(\frac{1}{z} \ln k\right)$

$\frac{1}{y^2} (\ln k)^2 = \frac{1}{xz} (\ln k)^2$

Assuming $\ln k \neq 0$ (i.e., $k \neq 1$), we can divide both sides by $(\ln k)^2$: $\frac{1}{y^2} = \frac{1}{xz}$

This simplifies to $y^2 = xz$. We know from the problem statement that $x$, $y$, and $z$ are in GP, which means $y^2 = xz$ is true. Therefore, Statement II is correct.

Conclusion

Based on the analysis, Statement I is incorrect because $p$, $q$, $r$ are in GP, not AP. Statement II is correct because the condition $y^2 = xz$ directly leads to $\ln p$, $\ln q$, $\ln r$ being in GP.

Thus, only Statement II is correct.