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Question Number 30156 by naybr last updated on 17/Feb/18

$$\mathrm{If}x,y,z\mathrm{are}\mathrm{in}\mathrm{GP}\mathrm{and}x+3,y+3,z+3\mathrm{are}$$$$\mathrm{in}\mathrm{HP},\mathrm{then}$$

Answered by ajfour last updated on 17/Feb/18

$$Then$$$$\frac{y}{x}=\frac{z}{y}=rand\frac{2}{y+3}=\frac{1}{x+3}+\frac{1}{z+3}$$$$letx+3=u,y+3=v,z+3=w$$$$\frac{2}{v}=\frac{u+w}{uw}and\frac{v-3}{u-3}=\frac{w-3}{v-3}=r$$$$\Rightarrow v(u+w)=2uwand$$$$v^2-6v+9=uw+9-3(u+w)$$$$\Rightarrow v^2-6v=uw-\frac{6uw}{v}$$$$\Rightarrow v^2(1-\frac{6}{v})=uw(1-\frac{6}{v})$$$$\Rightarrow \mathit{v}=6or\mathit{u},\mathit{v},\mathit{w}areinG.P.$$$$y+3=6orx+3,y+3,z+3$$$$areinG.P.aswell.$$$$\Rightarrow \mathit{y}=3or{(\mathit{y}+3)}^2=(\mathit{x}+3)(\mathit{z}+3).$$

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