a^{n} + b^{n}/ a^{n-1} + b^{n-1} = (a+b)/2 can we say n=1 by comparing exponents? is there any other solution of it?
$$ \frac{a^n + b^n}{a^{n-1}+b^{n-1}}=\frac{a+b}{2}$$ rearranging: $$a^n+b^n = ab^{n-1} + ba^{n-1}$$ By inspection, n=1 will satisfy the relation for all real a and b. Are there any other examples? ##n=2## ##a^2+b^2=2ab \Rightarrow (a-b)^2=0## ... so the relation with n=2 describes a curve (a parabola). So it kinda depends on what counts as an answer. (The problem is under-specified.)