Question

For three positive integers $p , q , r , x^{p q^2}=y^{y r}=z^{p^2 r }$ and $r = pq +1$ such that $3,3 \log _y x, 3 \log _z y$ $7 \log _x z$ are in AP with common difference $\frac{1}{2}$ Then $r-p-q$ is equal to

• A. 6
• B. 2
• C. 12
• D. $-6$

Answer 1

The Correct Option is B

Given: \[ \log_x y = \frac{q}{p}, \quad \log_y z = \frac{r}{q}, \quad \log_z x = \frac{p}{r}. \] The terms in A.P. are: \[ 3, \quad 3\log_x y, \quad 3\log_y z, \quad 7\log_z x. \] The common difference is: \[ \text{Common difference } = \frac{1}{2}. \] Equating differences: \[ 3\log_x y - 3 = \frac{1}{2}, \quad 3\log_y z - 3\log_x y = \frac{1}{2}. \] From the first equation: \[ 3\frac{q}{p} - 3 = \frac{1}{2} \implies \frac{q}{p} = \frac{7}{6}. \] From the second equation: \[ 3\frac{r}{q} - 3\frac{q}{p} = \frac{1}{2}. \] Substitute \( \frac{q}{p} = \frac{7}{6} \): \[ 3\frac{r}{q} - 3\cdot\frac{7}{6} = \frac{1}{2}. \] Simplify: \[ \frac{r}{q} = \frac{13}{6}. \] Thus: \[ r = \frac{13}{6}q, \quad r = pq + 1. \] Substitute: \[ pq = 6, \quad r = 7. \] Verify: \[ 3p^2 = 4q. \] Solve for \( p \) and \( q \): \[ p = 2, \quad q = 3, \quad r = 7. \] Finally: \[ r - p - q = 7 - 2 - 3 = 2. \] Thus, the answer is: \[ \boxed{2}. \]