Question

please is question ko btao ​

Answer 1

Given:

x² + 9y² + 25z² = xyz(15/x + 5/y + 3/z)

=> x² + 9y² + 25z² = 15yz + 5xz + 3xy

=> x² + 9y² + 25z² - 15yz - 5xz - 3xy = 0

=> (x)² + (3y)² + (5z²) - (3y)(5z) - (x)(5z) - (x)(3y) = 0

=> (1/2)[(x - 3y)² + (3y - 5z)² + (5z - x)²] = 0

=> x = 3y, 3y = 5z, x = 5z

⇒ x = 3y = 5z

=> x : y : z = 1 : (1/3) : (1/5)

Therefore,

x,y,z are in H.P

Hope it help!