If frac{a^{n+1} + b^{n+1}}{a^n + b^n} is the | vedclass

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(C) Given that $\frac{a^{n+1} + b^{n+1}}{a^n + b^n} = \frac{a+b}{2}$.Cross-multiplying,we get:$2(a^{n+1} + b^{n+1}) = (a+b)(a^n + b^n)$Expanding the r

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(C) Given that $\frac{a^{n+1} + b^{n+1}}{a^n + b^n} = \frac{a+b}{2}$.Cross-multiplying,we get:$2(a^{n+1} + b^{n+1}) = (a+b)(a^n + b^n)$Expanding the right side:$2a^{n+1} + 2b^{n+1} = a^{n+1} + ab^n + ba^n + b^{n+1}$Rearranging the terms:$a^{n+1} - ab^n - ba^n + b^{n+1} = 0$Factoring the expression:$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 $a^n = b^n$.Dividing by $b^n$ (assuming $b 
eq 0$):$(\frac{a}{b})^n = 1 = (\frac{a}{b})^0$Therefore,$n = 0$.

If $\frac{a^{n+1} + b^{n+1}}{a^n + b^n}$ is the $A.M.$ of $a$ and $b$,then $n=$

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