p, q and r are in AP, then prove that (p + 2q - r) (2q + r -p) (r + p - q) = 4pqr.

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Solution

Given $p, q, r$ are in $A P$
$∴ q - p = r - q$ (distance between $p$ & $q$ is equal to distance between $q$ & $r$ )
$∴ 2 q = r + p . . . . . . (1)$
$L H S = (p + 2 q - r) (2 q + r - p) (r + p - q)$
$= (p + r + p - r) (r + p + r - p) (r + p - q)$ [ $∵$ from eqn $(1)$ sub for $2 q$ ]
$= 2 p .2 r (2 q - q)$ (from eqn $(1)$ )
$= 4 p q r$
$= R H S$
Hence proved

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