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Question Number 101563 by harckinwunmy last updated on 03/Jul/20

Answered by bramlex last updated on 03/Jul/20

$$\left(2\right)\frac{1}{\mathrm{y}+\mathrm{z}},\frac{1}{\mathrm{z}+\mathrm{x}},\frac{1}{\mathrm{x}+\mathrm{y}}\to \mathrm{AP}$$$$\Rightarrow \frac{2}{\mathrm{z}+\mathrm{x}}=\frac{1}{\mathrm{y}+\mathrm{z}}+\frac{1}{\mathrm{x}+\mathrm{y}}$$$$\frac{2}{\mathrm{z}+\mathrm{x}}=\frac{\mathrm{x}+2\mathrm{y}+\mathrm{z}}{(\mathrm{y}+\mathrm{z})(\mathrm{x}+\mathrm{y})}$$$$2({\mathrm{y}}^2+\mathrm{yx}+\mathrm{zx}+\mathrm{zy})=(\mathrm{z}+\mathrm{x})(\mathrm{x}+2\mathrm{y}+\mathrm{z})$$$${2\mathrm{y}}^2+2\mathrm{yx}+2\mathrm{zx}+2\mathrm{zy}=$$$$\mathrm{zx}+2\mathrm{zy}+{\mathrm{z}}^2+{\mathrm{x}}^2+2\mathrm{xy}+\mathrm{xz}$$$${2\mathrm{y}}^2={\mathrm{x}}^2+{\mathrm{z}}^2\Rightarrow \mathrm{so}{\mathrm{x}}^2,{\mathrm{y}}^2,{\mathrm{z}}^2\to \mathrm{AP}$$

Answered by PRITHWISH SEN 2 last updated on 03/Jul/20

$$\because \mathrm{a},\mathrm{b}\mathrm{and}\mathrm{c}\mathrm{are}\mathrm{the}\mathrm{three}\mathrm{cosecutive}\mathrm{terms}\mathrm{of}$$$$\mathrm{a}\mathbf{linear}\mathbf{sequence}$$$$\therefore \mathrm{a}+\mathrm{c}=2\mathrm{b}$$$$\mathrm{now}\mathrm{a}(\frac{1}{\mathrm{b}}+\frac{1}{\mathrm{c}})+\mathrm{c}(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}})$$$$=\frac{\mathrm{b}({\mathrm{a}}^2+{\mathrm{c}}^2)+\mathrm{ac}(\mathrm{a}+\mathrm{c})}{\mathrm{abc}}$$$$=\frac{\mathrm{b}({\mathrm{a}}^2+{\mathrm{c}}^2)+2\mathrm{abc}}{\mathrm{abc}}=\frac{{(\mathrm{a}+\mathrm{c})}^2}{\mathrm{ac}}=\frac{2\mathrm{b}(\mathrm{a}+\mathrm{c})}{\mathrm{ac}}$$$$=2\mathrm{b}(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{c}})$$$$\therefore \mathrm{the}\mathrm{terms}\mathrm{are}\mathrm{also}\mathrm{the}3\mathrm{consecutive}\mathrm{terms}\mathrm{of}$$$$\mathrm{the}\mathrm{same}\mathrm{linear}\mathrm{sequence}\mathbf{proved}$$

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