Given,
apβ=a+(pβ1)d
aqβ=a+(qβ1)d
arβ=a+(rβ1)d
asβ=a+(sβ1)d
apβ,aqβ,arβ,asβ are in G.P
βapβaqββ=arβasββ
consider,
apβaqββ=aqβarββ
subtracting 1 on both sides, we get,
apβaqβββ1=aqβarβββ1
arββaqβaqββapββ=aqβapββ
a+(rβ1)dβ[a+(qβ1)d]a+(qβ1)dβ[a+(pβ1)d]β=aqβapββ
pβqqβrβ=aqβapββ............(1)
now consider,
aqβarββ=arβasββ
subtracting 1 on both sides, we get,
aqβarβββ1=arβasβββ1
aqββarβarββasββ=aqβarββ
a+(qβ1)dβ[a+(rβ1)d]a+(rβ1)dβ[a+(sβ1)d]β=aqβarββ
qβrrβsβ=aqβarββ
From (1)
qβrrβsβ=apβaqββ............(2)
From (1) and (2), we have,
aqβapββ=apβaqββ
Therefore (pβq),(qβr),(rβs) are in G.P