Question:
If a, b, c are in AP, show that
$(a+2 b-c)(2 b+c-a)(c+a-b)=4 a b c$
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