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

(N/A) Let the $A.P.$ be with first term $a$ and common difference $d$.
The terms are $a_p = a + (p-1)d, a_q = a + (q-1)d, a_r = a + (r-1)d, a_s = a + (s-1)d$.
Since $a_p, a_q, a_r, a_s$ are in $G.P.,$ the common ratio is $k = \frac{a_q}{a_p} = \frac{a_r}{a_q} = \frac{a_s}{a_r}$.
Using the property of ratios,$\frac{a_q - a_r}{a_p - a_q} = \frac{a_q(1 - k)}{a_p(1 - k)} = \frac{a_q}{a_p} = k$.
Also,$a_q - a_r = (a + (q-1)d) - (a + (r-1)d) = (q-r)d$ and $a_p - a_q = (p-q)d$.
Thus,$k = \frac{(q-r)d}{(p-q)d} = \frac{q-r}{p-q}$.
Similarly,$\frac{a_r - a_s}{a_q - a_r} = \frac{a_r(1 - k)}{a_q(1 - k)} = \frac{a_r}{a_q} = k$.
Thus,$k = \frac{(r-s)d}{(q-r)d} = \frac{r-s}{q-r}$.
Since both ratios equal $k,$ we have $\frac{q-r}{p-q} = \frac{r-s}{q-r},$ which implies $(p-q), (q-r), (r-s)$ are in $G.P.$