If `(a^(n)+b^n)/ (a^(n-1) +b^(n-1))` is the A.M. between a and b, then find the value of n.

Answer

n=1

Explanation

According to the given condition, we have
`(a+b)/2=(a^(n)+b^(n))/(a^(n-1)+b^(n-1))`
or `(a+b)(a^(n-1)+b^(n-1))=2(a^(n)+b^(n))`
or `a^(n)+ab^(n-1)+ba^(n-1)+b^(n)=2a^(n)+2b^(n)`
or `ab^(n-1)+a^(n-1)b=a^(n)+b^(n)`
or `ab^(n-1)-b^(n)=a^(n)-a^(n-1)b`
or `b^(n-1)(a-b)=a^(n-1)(a-b)`
or `b^(n-1)=a^(n-1)`
or `(a/b)^(n-1)=1=(a/b)^(0)`
`rArrn-1=0`
or n=1