A) 0
B) 1
C) 2
D) 3
Correct Answer: A
Solution :
A.M. between a and \[b=\frac{a+b}{2}\] According to question, \[\frac{a+b}{2}=\frac{{{a}^{n+1}}+{{b}^{n+1}}}{{{a}^{n}}+{{b}^{n}}}\] \[\Rightarrow \]\[a{{a}^{n}}+b{{a}^{n}}+a{{b}^{n}}+b{{b}^{n}}=2{{a}^{n+1}}+2{{b}^{n+1}}\] \[\Rightarrow \]\[2{{a}^{n+1}}+2{{b}^{n+1}}-{{a}^{n+1}}-b{{a}^{n}}-a{{b}^{n}}-{{b}^{n+1}}=0\] \[\Rightarrow \]\[{{a}^{n+1}}+{{b}^{n+1}}-b{{a}^{n}}-a{{b}^{n}}=0\] \[\Rightarrow \]\[{{a}^{n}}(a-b)-{{b}^{n}}(a-b)=0\] \[\Rightarrow \] \[(a-b)\,({{a}^{n}}-{{b}^{n}})=0\] But \[a-b\ne 0\] \[\Rightarrow \] \[{{a}^{n}}-{{b}^{n}}=0\]\[\Rightarrow \]\[{{a}^{n}}={{b}^{n}}\] \[\Rightarrow \] \[{{\left( \frac{a}{b} \right)}^{n}}=1={{\left( \frac{a}{b} \right)}^{0}}\Rightarrow n=0\]
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