Given a, b, c are in A.P.
∴ b – a = c – b
2 b = a + c
Squaring on both sides,
(a + c)2 = (2b)2
a2 + c2 + 2 ac = 4b2
(a – c)2 + 2 ac + 2 ac = 4 b2
[a2 + c2 = (a – c)2 + 2 ac]
(a – c)2 = 4b2 – 4ac
(a – c)2 = 4(b2 – ac)
Hence it is proved.
Aliter: Given a, b, c are in A.P.
b – a = c – b
2 b = a + c
To prove, (a – c)2 = 4(b2 – ac)
L.H.S. = (a – c)2
= a2 + c2 – 2ab
= (a + c)2 – 2ac – 2ac
= (a + c)2 – 4ac
= (2b)2 – 4ac (2b = a + c)
= 4 b2 – 4ac = 4(b2 – ac) = R.H.S
∴ L.H.S = R.H.S., Hence proved.
