If` (x^p+y^p)/(x^(p-1)+y^(p-1))=(x+y)/2` be the A.M. between x and y then find the value of p.

Explanation

Given that ` (x^(p) + y^(p))/(x^(n-1) + y^(p-1)) = (x+ y)/(2)`
` rArr 2x^(p) + 2y^(p) = x^(p) + x^(p-1) y + xy^(p-1) + y^(p)`
` rArr x^(p) + y^(p) - x^(p-1) y - xy^(p-1) = 0 `
` rArr x^(p-1) (x-y) - y^(p-1) (x -y) = 0 `
` rArr (x-y) (x^(p-1) -y^(p-1)) = 0 `
Since ` x ne y , x -y ne 0 `
`therefore x^(p-1) - y^(p-1) = 0 `
` rArr x^(p-1) = y^(p-1) `
` rArr ((x)/(y))^(p-1) = 1 = ((x)/(y))^(0)`
` therefore p - 1 = 0 `
` rArr p = 1 `