If $p^{th},q^{th},r^{th}$ and $s^{th}$ terms of an $A.P$ are in $G.P$, then show that  $(p-q)(q-r)(r-s)$ are also in $G.P$.

Given,

$a_p=a+(p-1)d$

$a_q=a+(q-1)d$

$a_r=a+(r-1)d$

$a_s=a+(s-1)d$

$a_p,a_q,a_r,a_s$ are in G.P

$\Rightarrow \frac{a_q}{a_p}=\frac{a_s}{a_r}$

consider,

$\frac{a_q}{a_p}=\frac{a_r}{a_q}$

subtracting 1 on both sides, we get,

$\frac{a_q}{a_p}-1=\frac{a_r}{a_q}-1$

$\frac{a_q-a_p}{a_r-a_q}=\frac{a_p}{a_q}$

$\frac{a+(q-1)d-[a+(p-1)d]}{a+(r-1)d-[a+(q-1)d]}=\frac{a_p}{a_q}$

$\frac{q-r}{p-q}=\frac{a_p}{a_q}$............(1)

now consider,

$\frac{a_r}{a_q}=\frac{a_s}{a_r}$

subtracting 1 on both sides, we get,

$\frac{a_r}{a_q}-1=\frac{a_s}{a_r}-1$

$\frac{a_r-a_s}{a_q-a_r}=\frac{a_r}{a_q}$

$\frac{a+(r-1)d-[a+(s-1)d]}{a+(q-1)d-[a+(r-1)d]}=\frac{a_r}{a_q}$

$\frac{r-s}{q-r}=\frac{a_r}{a_q}$

From (1)

$\frac{r-s}{q-r}=\frac{a_q}{a_p}$............(2)

From (1) and (2), we have,

$\frac{a_p}{a_q}=\frac{a_q}{a_p}$

Therefore $(p-q),(q-r),(r-s)$ are in G.P