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Question -

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

(ii) a2┬а+ c2┬а+ 4ac = 2 (ab + bc + ca)

(iii) a3┬а+ c3┬а+ 6abc = 8b3



Answer -

(i)┬а(a тАУ c)2┬а=4 (a тАУ b) (b тАУ c)

Let us expand theabove expression

a2┬а+ c2┬атАУ2ac = 4(ab тАУ ac тАУ b2┬а+ bc)

a2┬а+4c2b2┬а+ 2ac тАУ 4ab тАУ 4bc = 0

(a + c тАУ 2b)2┬а=0

a + c тАУ 2b = 0

Since a, b, c are inAP

b тАУ a = c тАУ b

a + c тАУ 2b = 0

a + c = 2b

Hence, (a тАУ c)2┬а=4 (a тАУ b) (b тАУ c)

(ii)┬аa2┬а+ c2┬а+4ac = 2 (ab + bc + ca)

Let us expand theabove expression

a2┬а+ c2┬а+4ac = 2 (ab + bc + ca)

a2┬а+ c2┬а+2ac тАУ 2ab тАУ 2bc = 0

(a + c тАУ b)2┬атАУb2┬а= 0

a + c тАУ b = b

a + c тАУ 2b = 0

2b = a+c

b = (a+c)/2

Since a, b, c are inAP

b тАУ a = c тАУ b

b = (a+c)/2

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

(iii)┬аa3┬а+ c3┬а+6abc = 8b3

Let us expand theabove expression

a3┬а+ c3┬а+6abc = 8b3

a3┬а+ c3┬атАУ(2b)3┬а+ 6abc = 0

a3┬а+(-2b)3┬а+ c3┬а+ 3a(-2b)c = 0

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

(a тАУ 2b + c)3┬а=0

a тАУ 2b + c = 0

a + c = 2b

b = (a+c)/2

Since a, b, c are inAP

a тАУ b = c тАУ b

b = (a+c)/2

Hence, a3┬а+c3┬а+ 6abc = 8b3

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