A.P !

If x x , z z and y y are in an arithmetic progression (in that order), where x x , y y and z z are distinct real numbers, evaluate x x z + z y z \dfrac{x}{x-z} + \dfrac{z}{y-z} .

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1 solution

Chew-Seong Cheong
Dec 25, 2017

Since x x , z z and y y are in an arithmetic progression, let x = z d x = z-d and (y = z+d), where d d is the common difference. Then we have:

x x z + z y z = z d d + z d = z d + 1 + z d = 1 \begin{aligned} \frac x{x-z} + \frac z{y-z} & = \frac {z-d}{-d} + \frac zd = - \frac zd + 1 + \frac zd = \boxed{1} \end{aligned}

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