Question

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

• A. 6
• B. 2
• C. 12
• D. $-6$

Answer 1

Answer: (B)

$p{q}^2={\log }_x\lambda$
$qr={\log }_y\lambda$
$p^{2}r={\log }_z\lambda$
${\log }_{y}x=\frac{qr}{p{q}^2}=\frac{r}{pq}\dots \dots .(1)$
${\log }_{x}z=\frac{p{q}^2}{p^{2}r}=\frac{q^2}{pr}\dots \dots \dots (2)$
${\log }_{z}y=\frac{p^{2}r}{qr}=\frac{p^2}{q}\dots \dots \dots (3)$
$3,\frac{3r}{pq},\frac{3p^2}{q},\frac{7q^2}{pr}\text{ in A.P }$
$\frac{3r}{pq}-3=\frac{1}{2}$
$r=\frac{7}{6}pq\dots \dots \dots (4)$
$r=pq+1$
$pq=6\dots \dots \dots (5)$
$r=7\dots \dots \dots \dots (6)$
$\frac{3p^2}{q}=4$
After solving $p=2$ and $q=3$