If $p,q,r$ are in $A.P.$ and $x,y,z$ are in $G.P.,$ then $x^{q-r}.y^{r-p}.z^{p-q}=$

The correct option is A.$1$

Let $d$ be the common difference of $A.P.$ and $r(\ne 0),$ the common ratio of $G.P,$

Since, $p,q,r$ are in AP

$\therefore q=p+d....\left(1\right)$

$r=p+2d$..(ii)$

Subtracting, $\left(ii\right)$ from $\left(i\right)$, we get,

$\Rightarrow q-r=-d....\left(iii\right)$

$\Rightarrow r-p=2d....\left(iv\right)$ [From $\left(ii\right)$]

$\Rightarrow p-q=-d....\left(v\right)$ [From $\left(i\right)$]

Also, $x,y,z$ are in GP with common ratio $r$

$\Rightarrow y=xr...\left(vi\right)$ and $z=x{r}^2....\left(vii\right)$

$\therefore x^{q-r}.y^{r-p}.z^{p-q}=x^{-d}\cdot (xr{)}^{2d}\cdot (x{r}^2{)}^{-d}$ [From $\left(iii\right),\left(iv\right),\left(v\right),\left(vi\right)$ and $\left(vii\right)$]

$=x^{-d+2d-d}\cdot r^{2d-2d}$

$=x^0\cdot r^0=1$

Hence,$x^{q-r}.y^{r-p}.z^{p-q}=1$