question_answer

The expression \[\frac{{{a}^{n+1}}+{{b}^{n+1}}}{{{a}^{n}}+{{b}^{n}}}\] is \[[a\ne b\ne 0]\] is (where a and b are unequal non-zero numbers)

A) A.M. between a and b if \[n=-1\]

B) G.M. between a and b if \[n=-\frac{1}{2}\]

C) H.M. between a and b if n = 0

D) All are correct

Correct Answer: B

Solution :

[b] Let  \[\frac{{{a}^{n+1}}+{{b}^{n+1}}}{{{a}^{n}}+{{b}^{n}}}=\frac{a+b}{2}\]

\[\Rightarrow \,\,\,2{{a}^{n+1}}+2{{b}^{n+1}}={{a}^{n+1}}+{{b}^{n+1}}+a{{b}^{n}}+b{{a}^{n}}\]

\[\Rightarrow \,\,(a-b)\,\,({{a}^{n}}-{{b}^{n}})=0\]

\[\Rightarrow \,\,{{a}^{n}}={{b}^{n}}\] it is possible for unequal numbers a

and b if \[n=0\]

Let \[\frac{{{a}^{n+1}}+{{b}^{n+1}}}{{{a}^{n}}+{{b}^{n}}}=\sqrt{ab}\]

\[\Rightarrow \,\,{{a}^{n+1}}+{{b}^{n+1}}={{a}^{n+\frac{1}{2}}}\,{{b}^{\frac{1}{2}}}+{{a}^{\frac{1}{2}}}\,{{b}^{n+\frac{1}{2}}}\]

\[\Rightarrow \,\,\,\left( {{a}^{n+\frac{1}{2}}}-{{b}^{n+\frac{1}{2}}} \right)\,\,\,\left( \sqrt{a}-\sqrt{b} \right)=0\]

\[\Rightarrow \,\,\,\,{{a}^{n+\frac{1}{2}}}-{{b}^{n+\frac{1}{2}}}=0,\]

which holds true if \[n+\frac{1}{2}=0\Rightarrow n=-\frac{1}{2}\]

Let \[\frac{{{a}^{n+1}}+{{b}^{n+1}}}{{{a}^{n}}+{{b}^{n}}}=\frac{2ab}{a+b}\]

\[\Rightarrow \,\,\,{{a}^{n+2}}+{{a}^{n+1}}b+a{{b}^{n+1}}+b_{=2a}^{n+2}\,{{\,}^{n+1}}b+2a{{b}^{n+1}}\]

\[\Rightarrow \,\,\,(a-b)\,\,({{a}^{n+1}}-{{b}^{n+1}})=0\]

\[\Rightarrow \,\,\,{{a}^{n+1}}-{{b}^{n+1}}=0\Rightarrow n=-1\]