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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.

$$\mathcal{I}{f}\:\:{a},{b},{c},{d}\:{are}\:{in}\:{G}.{P}.,\:{prove}\:{that} \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} ,{b}^{\mathrm{2}} −{c}^{\mathrm{2}} ,{c}^{\mathrm{2}} −{d}^{\mathrm{2}} \:{are}\:{also}\:{in}\:{G}.{P}. \\ $$

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.}

$${I}\:{assume}\:{that}\:{a},{b},{c},{d}\in\mathbb{C}−\left\{\mathrm{0}\right\}, \\ $$$${a},{b},{c},{d}\:{form}\:{a}\:{G}.{P}\:{in}\:{this}\:{sequence} \\ $$$${and}\:{a}\neq{b}\neq{c}\neq{d}.\:{If}\:{r}\:{is}\:{the}\:{common} \\ $$$${ratio}\:{of}\:{this}\:{G}.{P} \\ $$$$\Rightarrow\:\frac{{b}}{{a}}=\frac{{c}}{{b}}=\frac{{d}}{{c}}={r}\:\Rightarrow\:{ac}={b}^{\mathrm{2}} \:{and}\:{bd}={c}^{\mathrm{2}} . \\ $$$${Let}\:\psi=\frac{{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }.\: \\ $$$$\psi=\frac{{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }=\frac{{b}^{\mathrm{2}} −{bd}}{{a}^{\mathrm{2}} −{ac}}=\frac{{b}\left({b}−{d}\right)}{{a}\left({a}−{c}\right)}=\frac{{c}\left({b}−{d}\right)}{{b}\left({a}−{c}\right)} \\ $$$$\psi=\frac{{d}\left({b}−{d}\right)}{{c}\left({a}−{c}\right)}=\frac{{db}−{d}^{\mathrm{2}} }{{ac}−{c}^{\mathrm{2}} }=\frac{{c}^{\mathrm{2}} −{d}^{\mathrm{2}} }{{b}^{\mathrm{2}} −{c}^{\mathrm{2}} } \\ $$$${Since}\:\frac{{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }=\frac{{c}^{\mathrm{2}} −{d}^{\mathrm{2}} }{{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }=\psi={constant}, \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} ,\:{b}^{\mathrm{2}} −{c}^{\mathrm{2}} ,\:{c}^{\mathrm{2}} −{d}^{\mathrm{2}} \:{form}\:{a}\:{G}.{P}\:\:{in}\:{that} \\ $$$${sequence}. \\ $$$$\left\{\mathbb{C}\:{is}\:{the}\:{most}\:{general}\:{set}\:{of}\:{numbers}\right. \\ $$$${in}\:{current}\:{existence}\:{containing}\:{all} \\ $$$${conceivable}\:{numbers}.\:{This}\:{is}\:{why}\: \\ $$$$\mathbb{C}\:{was}\:{chosen}\:{as}\:{the}\:{domain}\:{for} \\ $$$${a},{b},{c}\:{and}\:{d}.\:{If}\:{any}\:{one}\:{of}\:{a},{b},{c}\:{or}\:{d}=\mathrm{0} \\ $$$$\left.{we}\:{have}\:{contradictions}\:{on}\:{r}.\right\} \\ $$

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