If a, b, c are in A.P., prove that

(b + c)² — a², (c + a)² — b², (a + b)² — c² are in A.P.

Given: a, b, c are in AP

Since, a, b, c are in AP, we have a + c = 2b …(i)

Now, (b + c)² – a², (c + a)² – b², (a + b)² – c² will be in A.P

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

i.e. if b + c – a, c + a – b, a + b – c are in AP

[dividing by (a + b + c)]

if (b + c – a) + (a + b – c) = 2(c + a – b)

if 2b = 2(c + a – b)

if b = c + a – b

if a + c = 2b which is true by (i)

Hence, (b + c)² – a², (c + a)² – b², (a + b)² – c² are in A.P