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a-b-c-are-in-G-P-If-a-x-b-y-c-z-prove-that-1-x-1-z-are-in-A-P-




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

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