Mathematics - Sequence & Series Question with Solution | TestHub

Let

$x^{p{q}^2}=y^{qr}=z^{p^{2}r}=\lambda$

Then,

$p{q}^2={\log }_x\lambda$

$qr={\log }_y\lambda$

$p^{2}r={\log }_z\lambda$

So,

$\frac{{\log }_y\lambda }{{\log }_x\lambda }=\frac{qr}{p{q}^2}$

$\Rightarrow {\log }_{y}x=\frac{r}{pq} ...\left(1\right)$

Similarly,

${\log }_{x}z=\frac{p{q}^2}{p^{2}r}=\frac{q^2}{pr} ....\left(2\right)$

${\log }_{z}y=\frac{p^{2}r}{qr}=\frac{p^2}{q} ...\left(3\right)$

So, $3, 3{\log }_{y}x, 3 {\log }_{z}y, 7{\log }_{x}z$ are in A.P., hence $3,\frac{3r}{pq},\frac{3p^2}{q},\frac{7q^2}{pr}$ are in A.P. Therefore,

$\frac{3r}{pq}-3=\frac{1}{2}$

$r=\frac{7}{6}pq$

Also,

$r=pq+1$

So,

$pq+1=\frac{7}{6}pq$

$\Rightarrow pq=6$

$\Rightarrow r=7$

So,

$\frac{3p^2}{q}=3+2\left(\frac{1}{2}\right)=4$

$\Rightarrow \frac{3p^3}{pq}=4$

$\Rightarrow p^3=8$

$\Rightarrow p=2$

$\Rightarrow q=3$

Hence, $r-p-q=7-2-3=2$