Suppose a is a fixed real number such that `(a - x)/(px) = (a - y)/(qy) = (a - z)/(rz)`
If p,q,r, are in A.P., then prove that x,y,z are in H.P.

Explanation

`:.` p, a, r are in A.P
`:. Q - p = r - q`...(i)
`rArr p - q = q - r = k` (let)
Given `(a - x)/(px) = (a - y)/(qy) = (a - z)/(rz) " "rArr" " ((a)/(x) - 1)/(p) = ((a)/(y) - 1)/(q) = ((a)/(z) - 1)/(r)`
`rArr (((a)/(x) - 1) - ((a)/(y) - 1))/(p - q) = (((a)/(y) - 1)- ((a)/(z) - 1))/(q - r)` (by law of proportion)
`rArr ((a)/(x) - (a)/(y))/(k) = ((a)/(y) - (a)/(z))/(k)` { from (i)}
`rArr a ((1)/(x) - (1)/(y)) = a ((1)/(y) - (1)/(z)) rArr " " (1)/(x) - (1)/(y) = (1)/(y) - (1)/(z)`
`:. (2)/(y) = (1)/(x) + (1)/(z)`
`:. (1)/(x), (1)/(y), (1)/(z)` are in A.P
Hence x, y, z are in H.P.