Question Number 17440 by virus last updated on 05/Jul/17
$$\mathrm{3}{x}−\mathrm{4}{y}=\mathrm{0},\mathrm{4}{y}−\mathrm{5}{z}=\mathrm{0},\mathrm{5}{z}−\mathrm{3}{x}=\mathrm{0} \\ $$$${then}\:{x},{y},{z}\:{is}\:{AP},{GP},{HP},{AGP}?????? \\ $$
Answered by sushmitak last updated on 06/Jul/17
$$\frac{{x}}{{y}}=\frac{\mathrm{4}}{\mathrm{3}}={k},\frac{{y}}{{z}}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${x}=\mathrm{4}{k} \\ $$$${y}=\mathrm{3}{k} \\ $$$${z}=\frac{\mathrm{12}{k}}{\mathrm{5}} \\ $$$${y}^{\mathrm{2}} =\mathrm{9}{k}\neq{xz}\:,{xyz}\:{are}\:{not}\:{in}\:{GP} \\ $$$$\mathrm{2}{y}=\mathrm{6}{k}\neq{x}+{z},{xyz}\:{are}\:{not}\:{in}\:{AP} \\ $$$$\frac{\mathrm{2}{xz}}{{x}+{z}}=\frac{\mathrm{2}\centerdot\mathrm{3}{k}\centerdot\frac{\mathrm{12}{k}}{\mathrm{5}}}{\mathrm{4}{k}+\frac{\mathrm{12}{k}}{\mathrm{5}}}=\frac{\mathrm{72}{k}}{\mathrm{32}}\neq{y}\:\:{x},{y},{z}\:{are}\:{not}\:{in}\:{HP} \\ $$