Question Number 144271 by 7770 last updated on 24/Jun/21
$${a},{b},{c}\:{are}\:{in}\:{G}.{P}.\:{If}\:{a}^{{x}} ={b}^{{y}} ={c}^{{z}} \\ $$$${prove}\:{that}\:\frac{\mathrm{1}}{{x}},\frac{\mathrm{1}}{{z}}\:{are}\:{in}\:{A}.{P}. \\ $$
Commented by Rasheed.Sindhi last updated on 24/Jun/21
$${Correct}\:{please}: \\ $$$${prove}\:{that}\:\frac{\mathrm{1}}{{x}},\frac{\mathrm{1}}{{y}},\frac{\mathrm{1}}{{z}}\:{are}\:{in}\:{A}.{P}. \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 24/Jun/21
$${Let}\:{a}^{{x}} ={b}^{{y}} ={c}^{{z}} ={k} \\ $$$$\:\:\:\:\:\:\:\:{x}=\frac{\mathrm{log}\:{k}}{\mathrm{log}\:{a}}\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{\mathrm{log}\:{a}}{\mathrm{log}\:{k}} \\ $$$$\:\:\:\:\:\:\:\:{y}=\frac{\mathrm{log}\:{k}}{\mathrm{log}\:{b}}\Rightarrow\frac{\mathrm{1}}{{y}}=\frac{\mathrm{log}\:{b}}{\mathrm{log}\:{k}} \\ $$$$\:\:\:\:\:\:\:\:{x}=\frac{\mathrm{log}\:{k}}{\mathrm{log}\:{c}}\Rightarrow\frac{\mathrm{1}}{{z}}=\frac{\mathrm{log}\:{c}}{\mathrm{log}\:{k}} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{{y}}−\frac{\mathrm{1}}{{x}}\overset{?} {=}\frac{\mathrm{1}}{{z}}−\frac{\mathrm{1}}{{y}} \\ $$$$\frac{\mathrm{log}\:{b}}{\mathrm{log}\:{k}}−\frac{\mathrm{log}\:{a}}{\mathrm{log}\:{k}}\overset{?} {=}\frac{\mathrm{log}\:{c}}{\mathrm{log}\:{k}}−\frac{\mathrm{log}\:{b}}{\mathrm{log}\:{k}} \\ $$$$\because\:{a},{b},{c}\:{are}\:{in}\:{GP} \\ $$$$\therefore\:{Let}\:{b}={ar}\:\&\:{c}={ar}^{\mathrm{2}} \\ $$$$\frac{\mathrm{log}\left({ar}\right)}{\mathrm{log}\:{k}}−\frac{\mathrm{log}\:{a}}{\mathrm{log}\:{k}}\overset{?} {=}\frac{\mathrm{log}\left({ar}^{\mathrm{2}} \right)}{\mathrm{log}\:{k}}−\frac{\mathrm{log}\left({ar}\right)}{\mathrm{log}\:{k}} \\ $$$$\frac{\mathrm{log}\:{a}+\mathrm{log}\:{r}−\mathrm{log}\:{a}}{\mathrm{log}\:{k}}\overset{?} {=}\frac{\mathrm{log}\:{a}+\mathrm{2log}\:{r}−\mathrm{log}\:{a}−\mathrm{log}\:{r}}{\mathrm{log}\:{k}} \\ $$$$\frac{\mathrm{log}\:{r}}{\mathrm{log}\:{k}}=\frac{\mathrm{log}\:{r}}{\mathrm{log}\:{k}}\: \\ $$$$\therefore\:\frac{\mathrm{1}}{{x}},\frac{\mathrm{1}}{{y}},\frac{\mathrm{1}}{{z}}\:{are}\:{in}\:{AP} \\ $$
Answered by Rasheed.Sindhi last updated on 24/Jun/21
$${Let}\:{b}={ar}\:\&\:{c}={ar}^{\mathrm{2}} \\ $$$${a}^{{x}} ={b}^{{y}} ={c}^{{z}} \Rightarrow{a}^{{x}} =\left({ar}\right)^{{y}} =\left({ar}^{\mathrm{2}} \right)^{{z}} \\ $$$${x}\mathrm{log}\:{a}={y}\left(\mathrm{log}\:{a}+\mathrm{log}\:{r}\right)={z}\left(\mathrm{log}\:{a}+\mathrm{2log}\:{r}\right)={k}\left({say}\right) \\ $$$${x}=\frac{{k}}{\mathrm{log}\:{a}}\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{\mathrm{log}\:{a}}{{k}} \\ $$$${y}=\frac{{k}}{\mathrm{log}\:{a}+\mathrm{log}\:{r}}\Rightarrow\frac{\mathrm{1}}{{y}}=\frac{\left.\mathrm{log}\:{a}+\mathrm{log}\:{r}\right)}{{k}} \\ $$$${z}=\frac{{k}}{\mathrm{log}\:{a}+\mathrm{2log}\:{r}}\Rightarrow\frac{\mathrm{1}}{{z}}=\frac{\mathrm{log}\:{a}+\mathrm{2log}\:{r}}{{k}} \\ $$$$\frac{\mathrm{1}}{{y}}−\frac{\mathrm{1}}{{x}}\overset{?} {=}\frac{\mathrm{1}}{{z}}−\frac{\mathrm{1}}{{y}} \\ $$$$\frac{\left.\mathrm{log}\:{a}+\mathrm{log}\:{r}\right)}{{k}}−\frac{\mathrm{log}\:{a}}{{k}}\overset{?} {=}\frac{\mathrm{log}\:{a}+\mathrm{2log}\:{r}}{{k}}−\frac{\left.\mathrm{log}\:{a}+\mathrm{log}\:{r}\right)}{{k}} \\ $$$$\frac{\mathrm{log}\:{r}}{{k}}=\frac{\mathrm{log}\:{r}}{{k}} \\ $$$$\therefore\:\frac{\mathrm{1}}{{x}},\frac{\mathrm{1}}{{y}},\frac{\mathrm{1}}{{z}}\:{are}\:{in}\:{AP} \\ $$
Answered by Rasheed.Sindhi last updated on 24/Jun/21
$${a}^{{x}} ={b}^{{y}} ={c}^{{z}} \\ $$$${a}={b}^{{y}/{x}} ={c}^{{z}/{x}} \\ $$$${b}={ar}\:\&\:{c}={ar}^{\mathrm{2}} \\ $$$${a}={a}^{{y}/{x}} {r}^{{y}/{x}} ={a}^{{z}/{x}} {r}^{\mathrm{2}{z}/{x}} \\ $$$${r}^{{y}/{x}} ={a}^{\mathrm{1}−\frac{{y}}{{x}}} ={a}^{\frac{{x}−{y}}{{x}}} \\ $$$${r}={a}^{\frac{{x}−{y}}{{y}}} …………..\left({i}\right) \\ $$$${a}={a}^{{z}/{x}} {r}^{\mathrm{2}{z}/{x}} \\ $$$${r}^{\mathrm{2}{z}/{x}} ={a}^{\mathrm{1}−\frac{{z}}{{x}}} ={a}^{\frac{{x}−{z}}{{x}}} \\ $$$${r}={a}^{\frac{{x}−{z}}{\mathrm{2}{z}}} ………………\left({ii}\right) \\ $$$${From}\:\left({i}\right)\:\&\:\left({ii}\right) \\ $$$${r}={a}^{\frac{{x}−{y}}{{y}}} ={a}^{\frac{{x}−{z}}{\mathrm{2}{z}}} \Rightarrow\frac{{x}−{y}}{{y}}=\frac{{x}−{z}}{\mathrm{2}{z}} \\ $$$$\mathrm{2}{xz}−\mathrm{2}{yz}={xy}−{yz} \\ $$$$\mathrm{2}{xz}={xy}+{yz} \\ $$$${Dividing}\:{by}\:{xyz}: \\ $$$$\frac{\mathrm{2}}{{y}}=\frac{\mathrm{1}}{{z}}+\frac{\mathrm{1}}{{x}} \\ $$$$\frac{\mathrm{1}}{{y}}=\frac{\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{z}}}{\mathrm{2}} \\ $$$$\therefore\:\frac{\mathrm{1}}{{x}},\frac{\mathrm{1}}{{y}},\frac{\mathrm{1}}{{z}}\:{are}\:{inAP} \\ $$