Question -
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