If the Arithmetic Mean of a and b is frac{a^{ | vedclass

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(C) The Arithmetic Mean of $a$ and $b$ is given by $\frac{a+b}{2}$.Given,$\frac{a^{n+1} + b^{n+1}}{a^n + b^n} = \frac{a+b}{2}$.Cross-multiplying,we ge

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(C) The Arithmetic Mean of $a$ and $b$ is given by $\frac{a+b}{2}$.Given,$\frac{a^{n+1} + b^{n+1}}{a^n + b^n} = \frac{a+b}{2}$.Cross-multiplying,we get $2(a^{n+1} + b^{n+1}) = (a+b)(a^n + b^n)$.$2a^{n+1} + 2b^{n+1} = a^{n+1} + ab^n + ba^n + b^{n+1}$.Rearranging the terms: $a^{n+1} - ab^n - ba^n + b^{n+1} = 0$.$a^n(a - b) - b^n(a - b) = 0$.$(a^n - b^n)(a - b) = 0$.Since $a 
eq b$,we must have $a^n - b^n = 0$,which implies $a^n = b^n$.Thus,$(\frac{a}{b})^n = 1 = (\frac{a}{b})^0$.Therefore,$n = 0$.

If the Arithmetic Mean of $a$ and $b$ is $\frac{a^{n+1} + b^{n+1}}{a^n + b^n}$,then $n = \dots$

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