Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

$${For}\:{b}\neq\mathrm{0} \\ $$$$\:{a}^{{n}+\mathrm{1}} ={b}^{{n}+\mathrm{1}} \Rightarrow\frac{{a}^{{n}+\mathrm{1}} }{{b}^{{n}+\mathrm{1}} }=\mathrm{1}\:\Rightarrow\left(\:\frac{{a}}{{b}}\right)^{{n}+\mathrm{1}} =\left(\frac{{a}}{{b}}\right)^{\mathrm{0}} \\ $$$$\Rightarrow{n}+\mathrm{1}=\mathrm{0}\Rightarrow{n}=−\mathrm{1} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com