Show that where α, β and γ are in A.P.

Let

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ + R₃, we get

Given that α, β and γ are in an A.P. Using the definition of an arithmetic progression, we have

β – α = γ – β

⇒ β + β = γ + α

⇒ 2β = γ + α

∴ α + γ = 2β

By substituting this in the above equation to find Δ, we get

Taking 2 common from R₁, we get

Applying R₁→ R₁ – R₂, we get

∴ Δ = 0

Thus, when α, β and γ are in A.P.