To prove: (a + 2b – c)(2b + c – a)(c + a – b) = 4abc.

Given: a, b, c are in A.P.

Proof: Since a, b, c are in A.P.

⇒ 2b = a + c … (i)

Taking LHS = (a + 2b – c) (2b + c – a) (c + a – b)

Substituting the value of 2b from eqn. (i)

= (a + a + c – c) (a + c + c – a) (c + a – b)

= (2a) (2c) (c + a – b)

Substituting the value of (a + c) from eqn. (i)

= (2a) (2c) (2b – b)

= (2a) (2c) (b)

= 4abc

= RHS

Hence Proved