Show that where a,b,c are in A.P.

Let


Recall that the value of a determinant remains same if we apply the operation Ri Ri + kRj or Ci Ci + kCj.


Applying R1 R1 + R3, we get




Given that a, b and c are in an A.P. Using the definition of an arithmetic progression, we have


b – a = c – b


b + b = c + a


2b = c + a


a + c = 2b


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 a, b and c are in A.P.


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