If x, y, z are in GP, then using properties of determinants, show that
, where x ≠ y ≠ z and p is any real number.
Given: x, y, z are in G.P
To show: 
Proof:

C1→ C1 – pC2 – C3



Expanding by R3
Δ = (-p2y – 2py – z)(xz – y2) …(i)
Since, x, y, z are in G.P. This means
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⇒ y2 = xz
⇒ y2 – xz = 0
or xz – y2 = 0
Substituting the value in eq. (i), we get
Δ = (-p2y – 2py – z)(0)
⇒ Δ = 0
∴ LHS = RHS
Hence Proved
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