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Question Number 43602 by peter frank last updated on 12/Sep/18

$$\mathrm{If}a,b,c\mathrm{are}\mathrm{in}\mathrm{GP}\mathrm{and}{\log }_{c}a,{\log }_{b}c,{\log }_{a}b$$$$\mathrm{are}\mathrm{in}\mathrm{AP},\mathrm{then}\mathrm{the}\mathrm{common}\mathrm{difference}$$$$\mathrm{of}\mathrm{the}\mathrm{AP}\mathrm{is}$$

Answered by $@ty@m last updated on 12/Sep/18

$$Let\log {}_{c}a=x-y\Rightarrow a=c^{x-y}...\left(1\right)$$$$\log {}_{b}c=x\Rightarrow b^x=c...\left(2\right)$$$$\log {}_{a}b=x+y\Rightarrow a^{x+y}=b...\left(3\right)$$$$From\left(1\right)\mathrm{\&}\left(2\right),$$$$a=b^{x(x-y)}...\left(4\right)$$$$From\left(3\right)\mathrm{\&}\left(4\right),$$$$a=a^{(x+y)x(x-y)}$$$$\Rightarrow (x+y)x(x-y)=1...\left(5\right)$$$$\because a,b\mathrm{\&}careinG.P.$$$$\therefore b^2=ac$$$$\Rightarrow a^{2(x+y)}=a.b^{x}from\left(2\right)\mathrm{\&}\left(3\right),$$$$\Rightarrow a^{2(x+y)}=a.a^{x+y}$$$$\Rightarrow a^{x+y}=a$$$$\Rightarrow x+y=1$$$$\Rightarrow x=1-y$$$$Substitutingthisin\left(5\right)$$$$x(x-y)=1$$$$\Rightarrow (1-y)(1-2y)=1$$$$\Rightarrow 1-3y+2y^2=1$$$$\Rightarrow y(2y-3)=0$$$$\Rightarrow y=0\left(inadmissible\right)$$$$ory=\frac{3}{2}Ans.$$

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