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Question Number 3615 by Rasheed Soomro last updated on 16/Dec/15
Find the value of n so that  ((a^(n+1) +b^(n+1) )/(a^n +b^n ))  may become the H.M. between  a   and    b.
$${Find}\:{the}\:{value}\:{of}\:{n}\:{so}\:{that} \\ $$$$\frac{{a}^{{n}+\mathrm{1}} +{b}^{{n}+\mathrm{1}} }{{a}^{{n}} +{b}^{{n}} } \\ $$$${may}\:{become}\:{the}\:{H}.{M}.\:{between} \\ $$$${a}\:\:\:{and}\:\:\:\:{b}. \\ $$
Commented by prakash jain last updated on 16/Dec/15
n=−1  ((1+1)/((1/a)+(1/b))) = ((2ab)/(a+b))
$${n}=−\mathrm{1} \\ $$$$\frac{\mathrm{1}+\mathrm{1}}{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}}\:=\:\frac{\mathrm{2}{ab}}{{a}+{b}} \\ $$
Answered by Yozzii last updated on 16/Dec/15
H.M, h=(1/((1/2)((1/a)+(1/b))))=((2ab)/(a+b)).  (a,b>0)  If h=((a^(n+1) +b^(n+1) )/(a^n +b^n ))  ⇒ ((a^(n+1) +b^(n+1) )/(a^n +b^n ))=((2ab)/(a+b))  ⇒(a+b)(a^(n+1) +b^(n+1) )=2ab(a^n +b^n )  a^(n+2) +abb^n +aba^n +b^(n+2) =2aba^n +2abb^n   a^(n+2) +b^(n+2) =aba^n +abb^n   a^(n+2) −ba^(n+1) −ab^(n+1) +b^(n+2) =0  a^(n+1) (a−b)−b^(n+1) (a−b)=0  (a^(n+1) −b^(n+1) )(a−b)=0  ⇒a=b or a^(n+1) =b^(n+1)  (∗).  If a≠b, (∗) is true iff n=−1.
$${H}.{M},\:{h}=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)}=\frac{\mathrm{2}{ab}}{{a}+{b}}.\:\:\left({a},{b}>\mathrm{0}\right) \\ $$$${If}\:{h}=\frac{{a}^{{n}+\mathrm{1}} +{b}^{{n}+\mathrm{1}} }{{a}^{{n}} +{b}^{{n}} } \\ $$$$\Rightarrow\:\frac{{a}^{{n}+\mathrm{1}} +{b}^{{n}+\mathrm{1}} }{{a}^{{n}} +{b}^{{n}} }=\frac{\mathrm{2}{ab}}{{a}+{b}} \\ $$$$\Rightarrow\left({a}+{b}\right)\left({a}^{{n}+\mathrm{1}} +{b}^{{n}+\mathrm{1}} \right)=\mathrm{2}{ab}\left({a}^{{n}} +{b}^{{n}} \right) \\ $$$${a}^{{n}+\mathrm{2}} +{abb}^{{n}} +{aba}^{{n}} +{b}^{{n}+\mathrm{2}} =\mathrm{2}{aba}^{{n}} +\mathrm{2}{abb}^{{n}} \\ $$$${a}^{{n}+\mathrm{2}} +{b}^{{n}+\mathrm{2}} ={aba}^{{n}} +{abb}^{{n}} \\ $$$${a}^{{n}+\mathrm{2}} −{ba}^{{n}+\mathrm{1}} −{ab}^{{n}+\mathrm{1}} +{b}^{{n}+\mathrm{2}} =\mathrm{0} \\ $$$${a}^{{n}+\mathrm{1}} \left({a}−{b}\right)−{b}^{{n}+\mathrm{1}} \left({a}−{b}\right)=\mathrm{0} \\ $$$$\left({a}^{{n}+\mathrm{1}} −{b}^{{n}+\mathrm{1}} \right)\left({a}−{b}\right)=\mathrm{0} \\ $$$$\Rightarrow{a}={b}\:{or}\:{a}^{{n}+\mathrm{1}} ={b}^{{n}+\mathrm{1}} \:\left(\ast\right). \\ $$$${If}\:{a}\neq{b},\:\left(\ast\right)\:{is}\:{true}\:{iff}\:{n}=−\mathrm{1}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 17/Dec/15
For b≠0   a^(n+1) =b^(n+1) ⇒(a^(n+1) /b^(n+1) )=1 ⇒( (a/b))^(n+1) =((a/b))^0   ⇒n+1=0⇒n=−1
$${For}\:{b}\neq\mathrm{0} \\ $$$$\:{a}^{{n}+\mathrm{1}} ={b}^{{n}+\mathrm{1}} \Rightarrow\frac{{a}^{{n}+\mathrm{1}} }{{b}^{{n}+\mathrm{1}} }=\mathrm{1}\:\Rightarrow\left(\:\frac{{a}}{{b}}\right)^{{n}+\mathrm{1}} =\left(\frac{{a}}{{b}}\right)^{\mathrm{0}} \\ $$$$\Rightarrow{n}+\mathrm{1}=\mathrm{0}\Rightarrow{n}=−\mathrm{1} \\ $$

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