Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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 (1/x),(1/z) are in A.P.

$${a},{b},{c}\:{are}\:{in}\:{G}.{P}.\:{If}\:{a}^{{x}} ={b}^{{y}} ={c}^{{z}} \\ $$$${prove}\:{that}\:\frac{\mathrm{1}}{{x}},\frac{\mathrm{1}}{{z}}\:{are}\:{in}\:{A}.{P}. \\ $$

Commented by Rasheed.Sindhi last updated on 24/Jun/21

Correct please: prove that (1/x),(1/y),(1/z) are in A.P.

$${Correct}\:{please}: \\ $$$${prove}\:{that}\:\frac{\mathrm{1}}{{x}},\frac{\mathrm{1}}{{y}},\frac{\mathrm{1}}{{z}}\:{are}\:{in}\:{A}.{P}. \\ $$$$ \\ $$

Answered by Rasheed.Sindhi last updated on 24/Jun/21

Let a^x =b^y =c^z =k x=((log k)/(log a))⇒(1/x)=((log a)/(log k)) y=((log k)/(log b))⇒(1/y)=((log b)/(log k)) x=((log k)/(log c))⇒(1/z)=((log c)/(log k)) (1/y)−(1/x)=^(?) (1/z)−(1/y) ((log b)/(log k))−((log a)/(log k))=^(?) ((log c)/(log k))−((log b)/(log k)) ∵ a,b,c are in GP ∴ Let b=ar & c=ar^2 ((log(ar))/(log k))−((log a)/(log k))=^(?) ((log(ar^2 ))/(log k))−((log(ar))/(log k)) ((log a+log r−log a)/(log k))=^(?) ((log a+2log r−log a−log r)/(log k)) ((log r)/(log k))=((log r)/(log k)) ∴ (1/x),(1/y),(1/z) are in AP

$${Let}\:{a}^{{x}} ={b}^{{y}} ={c}^{{z}} ={k} \\ $$$$\:\:\:\:\:\:\:\:{x}=\frac{\mathrm{log}\:{k}}{\mathrm{log}\:{a}}\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{\mathrm{log}\:{a}}{\mathrm{log}\:{k}} \\ $$$$\:\:\:\:\:\:\:\:{y}=\frac{\mathrm{log}\:{k}}{\mathrm{log}\:{b}}\Rightarrow\frac{\mathrm{1}}{{y}}=\frac{\mathrm{log}\:{b}}{\mathrm{log}\:{k}} \\ $$$$\:\:\:\:\:\:\:\:{x}=\frac{\mathrm{log}\:{k}}{\mathrm{log}\:{c}}\Rightarrow\frac{\mathrm{1}}{{z}}=\frac{\mathrm{log}\:{c}}{\mathrm{log}\:{k}} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{{y}}−\frac{\mathrm{1}}{{x}}\overset{?} {=}\frac{\mathrm{1}}{{z}}−\frac{\mathrm{1}}{{y}} \\ $$$$\frac{\mathrm{log}\:{b}}{\mathrm{log}\:{k}}−\frac{\mathrm{log}\:{a}}{\mathrm{log}\:{k}}\overset{?} {=}\frac{\mathrm{log}\:{c}}{\mathrm{log}\:{k}}−\frac{\mathrm{log}\:{b}}{\mathrm{log}\:{k}} \\ $$$$\because\:{a},{b},{c}\:{are}\:{in}\:{GP} \\ $$$$\therefore\:{Let}\:{b}={ar}\:\&\:{c}={ar}^{\mathrm{2}} \\ $$$$\frac{\mathrm{log}\left({ar}\right)}{\mathrm{log}\:{k}}−\frac{\mathrm{log}\:{a}}{\mathrm{log}\:{k}}\overset{?} {=}\frac{\mathrm{log}\left({ar}^{\mathrm{2}} \right)}{\mathrm{log}\:{k}}−\frac{\mathrm{log}\left({ar}\right)}{\mathrm{log}\:{k}} \\ $$$$\frac{\mathrm{log}\:{a}+\mathrm{log}\:{r}−\mathrm{log}\:{a}}{\mathrm{log}\:{k}}\overset{?} {=}\frac{\mathrm{log}\:{a}+\mathrm{2log}\:{r}−\mathrm{log}\:{a}−\mathrm{log}\:{r}}{\mathrm{log}\:{k}} \\ $$$$\frac{\mathrm{log}\:{r}}{\mathrm{log}\:{k}}=\frac{\mathrm{log}\:{r}}{\mathrm{log}\:{k}}\: \\ $$$$\therefore\:\frac{\mathrm{1}}{{x}},\frac{\mathrm{1}}{{y}},\frac{\mathrm{1}}{{z}}\:{are}\:{in}\:{AP} \\ $$

Answered by Rasheed.Sindhi last updated on 24/Jun/21

Let b=ar & c=ar^2 a^x =b^y =c^z ⇒a^x =(ar)^y =(ar^2 )^z xlog a=y(log a+log r)=z(log a+2log r)=k(say) x=(k/(log a))⇒(1/x)=((log a)/k) y=(k/(log a+log r))⇒(1/y)=((log a+log r))/k) z=(k/(log a+2log r))⇒(1/z)=((log a+2log r)/k) (1/y)−(1/x)=^(?) (1/z)−(1/y) ((log a+log r))/k)−((log a)/k)=^(?) ((log a+2log r)/k)−((log a+log r))/k) ((log r)/k)=((log r)/k) ∴ (1/x),(1/y),(1/z) are in AP

$${Let}\:{b}={ar}\:\&\:{c}={ar}^{\mathrm{2}} \\ $$$${a}^{{x}} ={b}^{{y}} ={c}^{{z}} \Rightarrow{a}^{{x}} =\left({ar}\right)^{{y}} =\left({ar}^{\mathrm{2}} \right)^{{z}} \\ $$$${x}\mathrm{log}\:{a}={y}\left(\mathrm{log}\:{a}+\mathrm{log}\:{r}\right)={z}\left(\mathrm{log}\:{a}+\mathrm{2log}\:{r}\right)={k}\left({say}\right) \\ $$$${x}=\frac{{k}}{\mathrm{log}\:{a}}\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{\mathrm{log}\:{a}}{{k}} \\ $$$${y}=\frac{{k}}{\mathrm{log}\:{a}+\mathrm{log}\:{r}}\Rightarrow\frac{\mathrm{1}}{{y}}=\frac{\left.\mathrm{log}\:{a}+\mathrm{log}\:{r}\right)}{{k}} \\ $$$${z}=\frac{{k}}{\mathrm{log}\:{a}+\mathrm{2log}\:{r}}\Rightarrow\frac{\mathrm{1}}{{z}}=\frac{\mathrm{log}\:{a}+\mathrm{2log}\:{r}}{{k}} \\ $$$$\frac{\mathrm{1}}{{y}}−\frac{\mathrm{1}}{{x}}\overset{?} {=}\frac{\mathrm{1}}{{z}}−\frac{\mathrm{1}}{{y}} \\ $$$$\frac{\left.\mathrm{log}\:{a}+\mathrm{log}\:{r}\right)}{{k}}−\frac{\mathrm{log}\:{a}}{{k}}\overset{?} {=}\frac{\mathrm{log}\:{a}+\mathrm{2log}\:{r}}{{k}}−\frac{\left.\mathrm{log}\:{a}+\mathrm{log}\:{r}\right)}{{k}} \\ $$$$\frac{\mathrm{log}\:{r}}{{k}}=\frac{\mathrm{log}\:{r}}{{k}} \\ $$$$\therefore\:\frac{\mathrm{1}}{{x}},\frac{\mathrm{1}}{{y}},\frac{\mathrm{1}}{{z}}\:{are}\:{in}\:{AP} \\ $$

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^2 a=a^(y/x) r^(y/x) =a^(z/x) r^(2z/x) r^(y/x) =a^(1−(y/x)) =a^((x−y)/x) r=a^((x−y)/y) ..............(i) a=a^(z/x) r^(2z/x) r^(2z/x) =a^(1−(z/x)) =a^((x−z)/x) r=a^((x−z)/(2z)) ..................(ii) From (i) & (ii) r=a^((x−y)/y) =a^((x−z)/(2z)) ⇒((x−y)/y)=((x−z)/(2z)) 2xz−2yz=xy−yz 2xz=xy+yz Dividing by xyz: (2/y)=(1/z)+(1/x) (1/y)=(((1/x)+(1/z))/2) ∴ (1/x),(1/y),(1/z) are inAP

$${a}^{{x}} ={b}^{{y}} ={c}^{{z}} \\ $$$${a}={b}^{{y}/{x}} ={c}^{{z}/{x}} \\ $$$${b}={ar}\:\&\:{c}={ar}^{\mathrm{2}} \\ $$$${a}={a}^{{y}/{x}} {r}^{{y}/{x}} ={a}^{{z}/{x}} {r}^{\mathrm{2}{z}/{x}} \\ $$$${r}^{{y}/{x}} ={a}^{\mathrm{1}−\frac{{y}}{{x}}} ={a}^{\frac{{x}−{y}}{{x}}} \\ $$$${r}={a}^{\frac{{x}−{y}}{{y}}} ..............\left({i}\right) \\ $$$${a}={a}^{{z}/{x}} {r}^{\mathrm{2}{z}/{x}} \\ $$$${r}^{\mathrm{2}{z}/{x}} ={a}^{\mathrm{1}−\frac{{z}}{{x}}} ={a}^{\frac{{x}−{z}}{{x}}} \\ $$$${r}={a}^{\frac{{x}−{z}}{\mathrm{2}{z}}} ..................\left({ii}\right) \\ $$$${From}\:\left({i}\right)\:\&\:\left({ii}\right) \\ $$$${r}={a}^{\frac{{x}−{y}}{{y}}} ={a}^{\frac{{x}−{z}}{\mathrm{2}{z}}} \Rightarrow\frac{{x}−{y}}{{y}}=\frac{{x}−{z}}{\mathrm{2}{z}} \\ $$$$\mathrm{2}{xz}−\mathrm{2}{yz}={xy}−{yz} \\ $$$$\mathrm{2}{xz}={xy}+{yz} \\ $$$${Dividing}\:{by}\:{xyz}: \\ $$$$\frac{\mathrm{2}}{{y}}=\frac{\mathrm{1}}{{z}}+\frac{\mathrm{1}}{{x}} \\ $$$$\frac{\mathrm{1}}{{y}}=\frac{\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{z}}}{\mathrm{2}} \\ $$$$\therefore\:\frac{\mathrm{1}}{{x}},\frac{\mathrm{1}}{{y}},\frac{\mathrm{1}}{{z}}\:{are}\:{inAP} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com