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Question
If the roots of (a – b)x2 + (b – c)x + (c – a) = 0 are real and equal, then prove that b, a, c are in arithmetic progression
Solution
(a – b)x2 + (b – c)x + (c – a) = 0
Here a = (a – b); b = b – c; c = c – a
Since the equation has real and equal roots ∆ = 0
∴ b2 – 4ac = 0
(b – c)2 – 4(a – b)(c – a) = 0
b2 + c2 – 2bc – 4 (ac – a2 – bc + ab) = 0
b2 + c2 – 2bc – 4ac + 4a2 + 4bc – 4ab = 0
b2 + c2 + 2bc – 4a (b + c) + 4a2 = 0
(b + c)2 – 4a (b + c) + 4a2 = 0
[(b+c) – 2a]2 = 0 ...[using a2 – 2ab + b2 = (a – b)2]
b + c – 2a = 0
b + c = 2a
b + c = a + a
c – a = a – b ...(t2 – t1 = t3 – t2)
b, a, c are in A.P.
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