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If-a-b-c-d-are-in-G-P-prove-that-a-2-b-2-b-2-c-2-c-2-d-2-are-also-in-G-P-

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Question Number 3464 by Rasheed Soomro last updated on 13/Dec/15
If a,b,c,d are in G.P., prove that a^2 −b^2 ,b^2 −c^2 ,c^2 −d^2 are also in G.P.
Ifa,b,c,dareinG.P.,provethata2−b2,b2−c2,c2−d2arealsoinG.P.
Answered by Yozzii last updated on 13/Dec/15
I assume that a,b,c,d∈C−{0}, a,b,c,d form a G.P in this sequence and a≠b≠c≠d. If r is the common ratio of this G.P ⇒ (b/a)=(c/b)=(d/c)=r ⇒ ac=b^2 and bd=c^2 . Let ψ=((b^2 −c^2 )/(a^2 −b^2 )). ψ=((b^2 −c^2 )/(a^2 −b^2 ))=((b^2 −bd)/(a^2 −ac))=((b(b−d))/(a(a−c)))=((c(b−d))/(b(a−c))) ψ=((d(b−d))/(c(a−c)))=((db−d^2 )/(ac−c^2 ))=((c^2 −d^2 )/(b^2 −c^2 )) Since ((b^2 −c^2 )/(a^2 −b^2 ))=((c^2 −d^2 )/(b^2 −c^2 ))=ψ=constant, a^2 −b^2 , b^2 −c^2 , c^2 −d^2 form a G.P in that sequence. {C is the most general set of numbers in current existence containing all conceivable numbers. This is why C was chosen as the domain for a,b,c and d. If any one of a,b,c or d=0 we have contradictions on r.}
Iassumethata,b,c,d∈C−{0},a,b,c,dformaG.Pinthissequenceanda≠b≠c≠d.IfristhecommonratioofthisG.P⇒ba=cb=dc=r⇒ac=b2andbd=c2.Letψ=b2−c2a2−b2.ψ=b2−c2a2−b2=b2−bda2−ac=b(b−d)a(a−c)=c(b−d)b(a−c)ψ=d(b−d)c(a−c)=db−d2ac−c2=c2−d2b2−c2Sinceb2−c2a2−b2=c2−d2b2−c2=ψ=constant,a2−b2,b2−c2,c2−d2formaG.Pinthatsequence.{Cisthemostgeneralsetofnumbersincurrentexistencecontainingallconceivablenumbers.ThisiswhyCwaschosenasthedomainfora,b,candd.Ifanyoneofa,b,cord=0wehavecontradictionsonr.}

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