The arithmetic mean of `Y` and `mu` and is `x^2 +1` and geometric mean of `Y` and `mu` is `2x + 1.` If harmonic mean of `Y` and `mu` is `4`, then `x` equals

Let x,y,z be different integers less than or equal to 14. The arithmetic mean of x and y is 10 and the geometric mean of x and z is 2 sqrt(21) then harmonic mean of y and z is equal to

The arithmetic mean of two numbers x and y is 3 and geometric mean is 1. Then x^(2)+y^(2) is equal to

The arithmetic mean of two numbers x and y is 3 and geometric mean is 1. Then x^(2) + y^(2) is equal to

The geometric mean of x and y is 6 and the geometric mean of x, y and z is also 6. Then the value of z is

The arithmetic mean of two numbers exceeds their geometric mean by 2 and the geometric mean exceeds their harmonic mean by 1.6. What are the two numbers ?

If x, y, z are in arithmetic progression and a is the arithmetic mean of x and y and b is the arithmetic mean of y and z, then prove that y is the arithmetic mean of a and b.

If x, y, z are in arithmetic progression and a is the arithmetic mean of x and y and b is the arithmetic mean of y and z, then prove that y is the arithmetic mean of a and b.

If x, y, z are in arithmetic progression and a is the arithmetic mean of x and y and b is the arithmetic mean of y and z, then prove that y is the arithmetic mean of a and b.

If the arithmetic mean of x and y is (x^n+y^n)/((x^(n-1)+y^(n-1))) ,then the value of n is