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

$I f a, b, c, d a r e i n G . P ., p r o v e t h a t$ $a^{2} - b^{2}, b^{2} - c^{2}, c^{2} - d^{2} a r e a l s o i n G . P .$
	Answered by Yozzii last updated on 13/Dec/15
	$I assume that a,b,c,d∈C−{0}, a,b,c,d form a G.P in this sequence and a≠b≠c≠d. If r is the common ratio of this G.P ⇒ (b/a)=(c/b)=(d/c)=r ⇒ ac=b^2 and bd=c^2 . Let ψ=((b^2 −c^2 )/(a^2 −b^2 )). ψ=((b^2 −c^2 )/(a^2 −b^2 ))=((b^2 −bd)/(a^2 −ac))=((b(b−d))/(a(a−c)))=((c(b−d))/(b(a−c))) ψ=((d(b−d))/(c(a−c)))=((db−d^2 )/(ac−c^2 ))=((c^2 −d^2 )/(b^2 −c^2 )) Since ((b^2 −c^2 )/(a^2 −b^2 ))=((c^2 −d^2 )/(b^2 −c^2 ))=ψ=constant, a^2 −b^2 , b^2 −c^2 , c^2 −d^2 form a G.P in that sequence. {C is the most general set of numbers in current existence containing all conceivable numbers. This is why C was chosen as the domain for a,b,c and d. If any one of a,b,c or d=0 we have contradictions on r.}$
	$I a s s u m e t h a t a, b, c, d \in C - {0},$ $a, b, c, d f o r m a G . P i n t h i s s e q u e n c e$ $a n d a \neq b \neq c \neq d . I f r i s t h e c o m m o n$ $r a t i o o f t h i s G . P$ $\Rightarrow \frac{b}{a} = \frac{c}{b} = \frac{d}{c} = r \Rightarrow a c = b^{2} a n d b d = c^{2} .$ $L e t ψ = \frac{b^{2} - c^{2}}{a^{2} - b^{2}} .$ $ψ = \frac{b^{2} - c^{2}}{a^{2} - b^{2}} = \frac{b^{2} - b d}{a^{2} - a c} = \frac{b (b - d)}{a (a - c)} = \frac{c (b - d)}{b (a - c)}$ $ψ = \frac{d (b - d)}{c (a - c)} = \frac{d b - d^{2}}{a c - c^{2}} = \frac{c^{2} - d^{2}}{b^{2} - c^{2}}$ $S i n c e \frac{b^{2} - c^{2}}{a^{2} - b^{2}} = \frac{c^{2} - d^{2}}{b^{2} - c^{2}} = ψ = c o n s t a n t,$ $a^{2} - b^{2}, b^{2} - c^{2}, c^{2} - d^{2} f o r m a G . P i n t h a t$ $s e q u e n c e .$ ${C i s t h e m o s t g e n e r a l s e t o f n u m b e r s$ $i n c u r r e n t e x i s t e n c e c o n t a i n i n g a l l$ $c o n c e i v a b l e n u m b e r s . T h i s i s w h y$ $C w a s c h o s e n a s t h e d o m a i n f o r$ $a, b, c a n d d . I f a n y o n e o f a, b, c o r d = 0$ $w e h a v e c o n t r a d i c t i o n s o n r .}$
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