Question Number 30156 by naybr last updated on 17/Feb/18
$$\mathrm{If}\:\:{x},\:{y},\:{z}\:\mathrm{are}\:\mathrm{in}\:\mathrm{GP}\:\mathrm{and}\:{x}+\mathrm{3},\:{y}+\mathrm{3},\:{z}+\mathrm{3}\:\mathrm{are} \\ $$$$\mathrm{in}\:\mathrm{HP},\:\mathrm{then} \\ $$
Answered by ajfour last updated on 17/Feb/18
$${Then} \\ $$$$\:\:\:\frac{{y}}{{x}}=\frac{{z}}{{y}}\:={r}\:\:\:{and}\:\:\:\frac{\mathrm{2}}{{y}+\mathrm{3}}=\frac{\mathrm{1}}{{x}+\mathrm{3}}+\frac{\mathrm{1}}{{z}+\mathrm{3}} \\ $$$${let}\:\:\:{x}+\mathrm{3}={u}\:,\:{y}+\mathrm{3}={v}\:,\:{z}+\mathrm{3}={w} \\ $$$$\frac{\mathrm{2}}{{v}}=\frac{{u}+{w}}{{uw}}\:\:\:\:{and}\:\:\:\frac{{v}−\mathrm{3}}{{u}−\mathrm{3}}=\frac{{w}−\mathrm{3}}{{v}−\mathrm{3}}\:={r} \\ $$$$\:\Rightarrow\:{v}\left({u}+{w}\right)=\mathrm{2}{uw}\:\:\:\:\:{and} \\ $$$${v}^{\mathrm{2}} −\mathrm{6}{v}+\mathrm{9}\:=\:{uw}+\mathrm{9}−\mathrm{3}\left({u}+{w}\right) \\ $$$$\Rightarrow\:\:\:{v}^{\mathrm{2}} −\mathrm{6}{v}={uw}−\frac{\mathrm{6}{uw}}{{v}} \\ $$$$\Rightarrow\:\:\:{v}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{6}}{{v}}\right)={uw}\left(\mathrm{1}−\frac{\mathrm{6}}{{v}}\right) \\ $$$$\Rightarrow\:\:\:\boldsymbol{{v}}=\mathrm{6}\:\:\:\:{or}\:\:\boldsymbol{{u}},\:\boldsymbol{{v}},\:\boldsymbol{{w}}\:\:{are}\:{in}\:{G}.{P}. \\ $$$$\:\:{y}+\mathrm{3}=\mathrm{6}\:\:\:\:{or}\:\:\:{x}+\mathrm{3}\:,\:{y}+\mathrm{3}\:,\:{z}+\mathrm{3}\: \\ $$$${are}\:{in}\:{G}.{P}.\:{as}\:{well}. \\ $$$$\Rightarrow\:\boldsymbol{{y}}=\mathrm{3}\:\:\:{or}\:\:\left(\boldsymbol{{y}}+\mathrm{3}\right)^{\mathrm{2}} =\left(\boldsymbol{{x}}+\mathrm{3}\right)\left(\boldsymbol{{z}}+\mathrm{3}\right)\:. \\ $$