(b + c – a), (c + a – b), (a + b – c) are in AP.

To prove: (b + c – a), (c + a – b), (a + b – c) are in AP.

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

Proof: Let d be the common difference for the A.P. a,b,c

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

⇒ b – a = c – b = common difference

⇒ a – b = b – c = d

⇒ 2(a – b) = 2(b – c) = 2d … (i)

Considering series (b + c – a), (c + a – b), (a + b – c)

For numbers to be in A.P. there must be a common difference between them

Taking (b + c – a) and (c + a – b)

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

= c + a – b – b – c + a

= 2a – 2b

= 2(a – b)

= 2d [from eqn. (i)]

Taking (c + a – b) and (a + b – c)

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

= a + b – c – c – a + b

= 2b – 2c

= 2(b – c)

= 2d [from eqn. (i)]

Here we can see that we have obtained a common difference between numbers i.e. 2d

Hence, (b + c – a), (c + a – b), (a + b – c) are in AP.