2009/008) If a,b,c are in AP then prove that (b+c)^2 - a^2 , (c+a)^2 - b^2 , (a+b)^2 - c^2 are in AP

let the terms be a1, a2,a3
a1 = (b+c)^2 - a^2 = (a+b+c)(b+c-a) = (a+b+c)(a+b+c - 2a)
a2 = (c+a)^2 - b^2 = (a+b+c)(a+c-b) = (a+b+c)(a+b+c - 2b)
a3 = (a+b)^2 - c^2 = (a+b+c)(a+b-c) = (a+b+c) (a+b+c - 2c)

as a b c are in ap so are -a , -b , -c so are -2a, -2b, - 2c

addiing constant to them we have

(a+b+c-2a), (a+b+c-2b), (a + b - 2c) are in AP

multiplying by (a+b+c) we have
(a+b+c)(a+b+c-2a), (a+b+c)(a+b+c-2b), (a+b+c) (a + b - 2c) are in AP

hence proved