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 \[\frac{{{a}^{n+1}}+{{b}^{n+1}}}{{{a}^{n}}+{{b}^{n}}}=\frac{2ab}{a+b}\] |
\[\Rightarrow \,\,\,{{a}^{n+2}}+{{a}^{n+1}}b+a{{b}^{n+1}}+b_{=2a}^{n+2}\,{{\,}^{n+1}}b+2a{{b}^{n+1}}\] |
\[\Rightarrow \,\,\,(a-b)\,\,({{a}^{n+1}}-{{b}^{n+1}})=0\] |
\[\Rightarrow \,\,\,{{a}^{n+1}}-{{b}^{n+1}}=0\Rightarrow n=-1\] |
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