please is question ko btao
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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!
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