AP in GP or GP in AP?

Let first term and common difference of this A.P. are $a$ and $d$ respectively.So,

$t_p=a+(p-1)d\,\,;t_q=a+(q-1)d\,\,;t_r=a+(r-1)d\,\,;t_s=a+(s-1)d$ .

According to the question $t_p,t_q,t_r,t_s$ form a G.P.Let common ratio is $m$

Then, $t_q t_r=t_p t_s \implies \dfrac{t_q}{t_p}=\dfrac{t_s}{t_r} \\ \implies \dfrac{a+(q-1)d}{a+(p-1)d}=\dfrac{a+(s-1)d}{a+(r-1)d}$

Using properties of fraction:-

$\dfrac{a+(q-1)d-a-(p-1)d}{a+(p-1)d}=\dfrac{a+(s-1)d-a-(r-1)d}{a+(r-1)d}$

$\dfrac{q-p}{a+(p-1)d}=\dfrac{s-r}{a+(r-1)d} \implies\dfrac{p-q}{a+(p-1)d}=\dfrac{r-s}{a+(r-1)d} \implies \dfrac{r-s}{p-q}=\dfrac{t_r}{t_p}=m^2\\ \implies (r-s)=m^2 (p-q)$

Again re-arrange the fraction : $\dfrac{t_q}{t_s}=\dfrac{t_p}{t_r} \implies \dfrac{t_q-t_s}{t_s}=\dfrac{t_p-t_r}{t_r} \implies \dfrac{q-s}{t_s}=\dfrac{p-r}{t_r} \\ \implies \dfrac{q-s}{p-r}=\dfrac{t_s}{t_r} \\ \implies \dfrac{q-s}{p-r}=m \\ \implies q-s=m(p-r) \\ \implies (q-s)=m(p-q+q-r) \\ \implies (q-r)+(r-s)=m(p-q)+m(q-r) \\ \implies (q-r) + m^2(p-q)=m(p-q)+m(q-r) ........\text{from the first derivation} \\ \implies (q-r)(m-1)=m(p-q)(m-1) \\ \implies (q-r)=m(p-q)..........\text{as m is not equal to 1}$

So, our final results are $(r-s)=m^2 (p-q)$ and $(q-r)=m(p-q)$

From above two results it is concluded that $p-q,q-r,r-s$ are in G.P.

This is not a correct solution but for objective point of view this is the best approach

Consider the AP To be natural numbers

you obtain p = 1 q=2 r = 4 s = 8

obviously -1,-2,-4 are IN GP AGAIN

Q.E.D