If a, b, c are in AP, show that

Solution:

To prove: $(a+2 b-c)(2 b+c-a)(c+a-b)=4 a b c$.

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

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

$\Rightarrow 2 b=a+c \ldots$ (i)

Taking $L H S=(a+2 b-c)(2 b+c-a)(c+a-b)$

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

$=(a+a+c-c)(a+c+c-a)(c+a-b)$

$=(2 a)(2 c)(c+a-b)$

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

$=(2 a)(2 c)(2 b-b)$

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

$=4 a b c$

= RHS

Hence Proved