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Question -

If a, b, c are in A.P., prove that:
(i) (a – c)2 = 4 (a – b) (b – c)

(ii) a2 + c2 + 4ac = 2 (ab + bc + ca)

(iii) a3 + c3 + 6abc = 8b3



Answer -

(i) (a – c)2 =4 (a – b) (b – c)

Let us expand theabove expression

a2 + c2 –2ac = 4(ab – ac – b2 + bc)

a2 +4c2b2 + 2ac – 4ab – 4bc = 0

(a + c – 2b)2 =0

a + c – 2b = 0

Since a, b, c are inAP

b – a = c – b

a + c – 2b = 0

a + c = 2b

Hence, (a – c)2 =4 (a – b) (b – c)

(ii) a2 + c2 +4ac = 2 (ab + bc + ca)

Let us expand theabove expression

a2 + c2 +4ac = 2 (ab + bc + ca)

a2 + c2 +2ac – 2ab – 2bc = 0

(a + c – b)2 –b2 = 0

a + c – b = b

a + c – 2b = 0

2b = a+c

b = (a+c)/2

Since a, b, c are inAP

b – a = c – b

b = (a+c)/2

Hence, a2 +c2 + 4ac = 2 (ab + bc + ca)

(iii) a3 + c3 +6abc = 8b3

Let us expand theabove expression

a3 + c3 +6abc = 8b3

a3 + c3 –(2b)3 + 6abc = 0

a3 +(-2b)3 + c3 + 3a(-2b)c = 0

Since, if a + b + c =0, a3 + b3 + c3 = 3abc

(a – 2b + c)3 =0

a – 2b + c = 0

a + c = 2b

b = (a+c)/2

Since a, b, c are inAP

a – b = c – b

b = (a+c)/2

Hence, a3 +c3 + 6abc = 8b3

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