In the case $s=p/q$ with $|p| \geq 3$, one can prove that there are is no $3$-term AP in $P_s$, P_s$. The proof is by reducing to the case$s$is an integer, s=p$, as follows.
Let $A=a^p$, $B=b^p$, $C=c^p$ such that $A^{1/q}+C^{1/q}=2B^{1/q} \quad (*) \quad$ and $(A,B,C)=1$. Let $K=\mathbf{Q}(\zeta_q)$ be the $q$-th cyclotomic field and $L=K(A^{1/q},B^{1/q},C^{1/q})$. Then $L/K$ is a finite abelian extension of exponent dividing $q$ and by Kummer theory, such extensions are in natural bijection with the finite subgroups of $K^{\times}/(K^{\times})^q$. The extension $L/K$ corresponds to the subgroup generated by the classes $\overline{A},\overline{B}, \overline{C}$ of $A,B,C$ in $K^{\times}/(K^{\times})^q$. In view of the following lemma, it suffices to prove $\overline{A}=\overline{B}=\overline{C}=1$.L=K$. Lemma 1. If$n^p$is a$q$-th power in$K$, then$n$is a$q$-th power in$\mathbf{Z}$. We also remark that Lemma 2. The integer$(A,B)=1$(any common B$ is relatively prime factor of to $A$ and to $B$ has C$. Proof. By symmetry, it suffices to divide prove$C$, by reasoning (A,B)=1$. Let $\ell$ be a prime number dividing $A$ and $B$. Then $\ell^{1/q}$ divides $A^{1/q}$ and $B^{1/q}$ in the ring $\overline{\mathbf{Z}}$ of all algebraic integersof $L$), and similarly . By $(C,B)=1$. It (*)$it follows that the subgroups$\langle \ell^{1/q} | C^{1/q}$. Thus$C/\ell \overline{A},\overline{C} in \rangle$and$\langle mathbf{Q} \overline{B} cap \rangle$intersect trivially in$K^{\times}/(K^{\times})^q$. In other words, the extensions$K(A^{1/q},C^{1/q})$and overline{\mathbf{Z}} = \mathbf{Z}$ which contradicts $K(B^{1/q})$ are linearly disjoint(A,B,C)=1$. This proves Lemma 2. If By equation$(*)$, we had have$B^{1/q} K(B^{1/q}) \not\in K$, then there would exist subset K(A^{1/q},C^{1/q})$ which reads $\sigma \overline{B} \in \mathrm{Gal}(L/K)$ fixing langle \overline{A},\overline{C} \rangle$in$A^{1/q},C^{1/q}$but not K^{\times}/(K^{\times})^q$. We can thus write $B^{1/q}$. By applying B \equiv A^{\alpha} C^{\gamma} \pmod{(K^{\times})^q}$for some$\sigma$\alpha,\gamma \geq 0$. By a reasoning similar to the identity $A^{1/q}+C^{1/q}=2B^{1/q}$, Lemma 1, we get a contradiction. Thus $B^{1/q} \in K$ and by the lemma deduce that $B$ B/(A^{\alpha} C^{\gamma})$is a$q$-th power in$\mathbf{Z}$, so that$A^{1/q}+C^{1/q} \\mathbf{Q}$but since this fraction is in \mathbf{Z}$.
By a similar reasoning lowest terms (Lemma 2), we get that $A/C$ B$is a$q$-th power in$\mathbf{Q}$, thus we can write$A=\lambda u^q$and$C=\lambda v^q$with \mathbf{Z}$.
Now let $\lambda,u,v \in \mathbf{Z}$. Replacing \sigma$be an aribtrary element in$(*)$we get \mathrm{Gal}(L/K)$. We have $\lambda^{1/q} \sigma(A^{1/q}) = \in zeta \mathbf{Q}$, so that cdot A^{1/q}$and$\lambda$is a \sigma(C^{1/q})=\zeta' \cdot C^{1/q}$ for some $q$-th power in roots of unity $\mathbf{Z}$ \zeta$and$\zeta'$. Considering the real parts of both sides of$\sigma(*)$, we are donesee that necessarily$\zeta=\zeta'=1$. This shows that$L=K$as requested. 2 edited body In the case$s=p/q$with$|p| \geq 3$, one can prove that there are no$3$-term AP in$P_s$, by reducing to the case$s$is an integer, as follows. Let$A=a^p$,$B=b^p$,$C=c^p$such that$A^{1/q}+C^{1/q}=2B^{1/q} \quad (*) \quad$and$(A,B,C)=1$. Let$K=\mathbf{Q}(\zeta_q)$be the$q$-th cyclotomic field and$L=K(A^{1/q},B^{1/q},C^{1/q})$. Then$L/K$is a finite abelian extension of exponent dividing$q$and by Kummer theory, such extensions are in natural bijection with the finite subgroups of$K^{\times}/(K^{\times})^q$. The extension$L/K$corresponds to the subgroup generated by the classes$\overline{A},\overline{B}, \overline{C}$of$A,B,C$in$K^{\times}/(K^{\times})^q$. In view of the following lemma, it suffices to prove$\overline{A}=\overline{B}=\overline{C}=1$. Lemma. If$n^p$is a$q$-th power in$K$, then$n$is a$q$-th power in$\mathbf{Z}$. Proof. Assume$n^p=\alpha^q$with$\alpha \in K$, then taking the norm we get$n^{p(q-1)}=N_{K/\mathbf{Q}}(\alpha)^q$. Since$\alpha$is an algebraic integer, we get that$n^{p(q-1)}$is a$q$-th power in$\mathbf{Z}$, and since$p(q-1)$and$q$are coprime, we get the result. We also remark that$(A,B)=1$(any common prime factor of$A$and$B$has to divide$C$, by reasoning in the ring of integers of$L$), and similarly$(C,B)=1$. It follows that the subgroups$\langle \overline{A},\overline{C} \rangle$and$\langle \overline{B} \rangle$intersect trivially in$K^{\times}/(K^{\times})^q$. In other words, the extensions$K(A^{1/q},C^{1/q})$and$K(B^{1/q})$are linearly disjoint. If we had$B^{1/q} \not\in K$, then there would exist$\sigma \in \mathrm{Gal}(L/K)$fixing$A^{1/q},C^{1/q}$but not$B^{1/q}$. By applying$\sigma$to the identity$A^{1/q}+C^{1/q}=2B^{1/q}$, we get a contradiction. Thus$B^{1/q} \in K$and by the lemma$B$is a$q$-th power in$\mathbf{Z}$, so that$A^{1/q}+C^{1/q} \in \mathbf{Z}$. By a similar reasoning$A/C$is a$q$-th power in$\mathbf{Q}$, thus we can write$A=\lambda u^q$and$C=\lambda v^q$with$\lambda,u,v \in \mathbf{Z}$. Replacing in$(*)$we get$\lambda^{p/q} \lambda^{1/q} \in \mathbf{Q}$, so that$\lambda$is a$q$-th power in$\mathbf{Z}$and we are done. 1 In the case$s=p/q$with$|p| \geq 3$, one can prove that there are no$3$-term AP in$P_s$, by reducing to the case$s$is an integer, as follows. Let$A=a^p$,$B=b^p$,$C=c^p$such that$A^{1/q}+C^{1/q}=2B^{1/q} \quad (*) \quad$and$(A,B,C)=1$. Let$K=\mathbf{Q}(\zeta_q)$be the$q$-th cyclotomic field and$L=K(A^{1/q},B^{1/q},C^{1/q})$. Then$L/K$is a finite abelian extension of exponent dividing$q$and by Kummer theory, such extensions are in natural bijection with the finite subgroups of$K^{\times}/(K^{\times})^q$. The extension$L/K$corresponds to the subgroup generated by the classes$\overline{A},\overline{B}, \overline{C}$of$A,B,C$in$K^{\times}/(K^{\times})^q$. In view of the following lemma, it suffices to prove$\overline{A}=\overline{B}=\overline{C}=1$. Lemma. If$n^p$is a$q$-th power in$K$, then$n$is a$q$-th power in$\mathbf{Z}$. Proof. Assume$n^p=\alpha^q$with$\alpha \in K$, then taking the norm we get$n^{p(q-1)}=N_{K/\mathbf{Q}}(\alpha)^q$. Since$\alpha$is an algebraic integer, we get that$n^{p(q-1)}$is a$q$-th power in$\mathbf{Z}$, and since$p(q-1)$and$q$are coprime, we get the result. We also remark that$(A,B)=1$(any common prime factor of$A$and$B$has to divide$C$, by reasoning in the ring of integers of$L$), and similarly$(C,B)=1$. It follows that the subgroups$\langle \overline{A},\overline{C} \rangle$and$\langle \overline{B} \rangle$intersect trivially in$K^{\times}/(K^{\times})^q$. In other words, the extensions$K(A^{1/q},C^{1/q})$and$K(B^{1/q})$are linearly disjoint. If we had$B^{1/q} \not\in K$, then there would exist$\sigma \in \mathrm{Gal}(L/K)$fixing$A^{1/q},C^{1/q}$but not$B^{1/q}$. By applying$\sigma$to the identity$A^{1/q}+C^{1/q}=2B^{1/q}$, we get a contradiction. Thus$B^{1/q} \in K$and by the lemma$B$is a$q$-th power in$\mathbf{Z}$, so that$A^{1/q}+C^{1/q} \in \mathbf{Z}$. By a similar reasoning$A/C$is a$q$-th power in$\mathbf{Q}$, thus we can write$A=\lambda u^q$and$C=\lambda v^q$with$\lambda,u,v \in \mathbf{Z}$. Replacing in$(*)$we get$\lambda^{p/q} \in \mathbf{Q}$, so that$\lambda$is a$q$-th power in$\mathbf{Z}\$ and we are done.