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Question Number 118298 by hmh2k20 last updated on 16/Oct/20

$$\mathrm{b},\mathrm{x},\mathrm{y},\mathrm{c}\mathrm{are}\mathrm{consecutive}\mathrm{terms}\mathrm{of}\mathrm{a}\mathrm{G}.\mathrm{P}.$$$$\therefore \mathbf{x}=\mathbf{br},\mathbf{y}={\mathbf{br}}^2,\mathbf{c}={\mathbf{br}}^3$$$$$$$$\mathrm{A}.\mathrm{M}.\mathrm{between}\mathrm{b}\mathrm{and}\mathrm{c}\mathrm{is}\mathrm{a}$$$$\therefore \frac{\mathrm{b}+\mathrm{c}}{2}=\mathrm{a}\Rightarrow 2\mathbf{a}=\mathbf{b}+\mathbf{c}$$$$$$$$2\mathbf{a}\mathrm{bc}=(\mathbf{b}+\mathbf{c})\mathrm{bc}$$$$={\mathrm{b}}^2\mathbf{c}+\mathrm{b}{\mathbf{c}}^2$$$$={\mathrm{b}}^2\left({\mathbf{br}}^3\right)+\mathrm{b}{\left({\mathbf{br}}^3\right)}^2$$$$={\mathrm{b}}^3{\mathrm{r}}^3+{\mathrm{b}}^3{\mathrm{r}}^6$$$$={\left(\mathbf{br}\right)}^3+{\left({\mathbf{br}}^2\right)}^3$$$$={\mathbf{x}}^3+{\mathbf{y}}^3$$$$$$$$\therefore {\mathrm{x}}^3+{\mathrm{y}}^3=2\mathrm{abc}\leftarrow$$

Commented by prakash jain last updated on 16/Oct/20

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