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Question Number 100570 by bshahid010@gmail.com last updated on 27/Jun/20

Commented by Dwaipayan Shikari last updated on 27/Jun/20

$${a}^{{x}} ={b}^{{y}} ={c}^{{z}} ={d}^{{w}} ={k}\:\:\left({k}\neq\mathrm{0}\right) \\$$$${a}={k}^{\frac{\mathrm{1}}{{x}}} \:\:{b}={k}^{\frac{\mathrm{1}}{{y}}} \:\:{c}={k}^{\frac{\mathrm{1}}{{z}}} \:\:{d}={k}^{\frac{\mathrm{1}}{{w}}} \\$$$$\frac{{a}}{{b}}=\frac{{b}}{{c}}=\frac{{c}}{{d}}\:\:\:\:\left({As}\:{they}\:{are}\:{in}\:{G}.{P}\right) \\$$$${ad}={bc}\:\:{so}\:\:\:\:\:\:{k}^{\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{w}}} ={k}^{\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}} \Rightarrow\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{w}}=\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}\:} \\$$$${or}\:\:\:\:{ac}={b}^{\mathrm{2}} \:\:\:\:\:\:{k}^{\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{z}}} ={k}^{\frac{\mathrm{2}}{{y}\:}} \:\:\:\Rightarrow\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{z}}=\frac{\mathrm{2}}{{y}} \\$$$${or}\:\:{c}^{\mathrm{2}} ={bd}\:\:\:\:{k}^{\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{w}}} ={k}^{\frac{\mathrm{2}}{{z}\:\:}} \:\:\Rightarrow\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{w}}=\frac{\mathrm{2}}{{z}} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{They}\:{are}\:{in}\:{H}.{P} \\$$