# are in A.P., show that are in A.P. provided a + b + c 0

Given: are in AP

Taking LCM

b2c + c2b + a2b + ab2 – 2ac2 – 2a2c = 0

b2c + c2b + a2b + ab2 –ac2 – ac2 – a2c – a2c = 0

(b2c – a2c) + (c2b – ac2) + (a2b – a2c) + (ab2 – ac2) = 0

c (b – a)(b + a) + c2(b – a) + a2 (b – c) + a(b + c)(b – c) = 0

c(b – a) {(b + a) + c} + a(b – c) {a + (b + c)} = 0

(a + b + c){cb – ca + ab – ca} = 0

Given a + b + c ≠ 0

cb – ca + ab – ca = 0

cb – 2ca + ab = 0

are in AP

Hence Proved

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