If a, b, c are in AP, prove that

(i) (a – c)2 = 4(a – b)(b – c)


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


(iii) a3 + c3 + 6abc = 8b3


(i) (a – c)2 = 4(a – b)(b – c)


To prove: (a – c)2 = 4(a – b)(b – c)


Given: a, b, c are in A.P.


Proof: Since a, b, c are in A.P.


c – b = b – a = common difference


b – c = a – b … (i)


And, 2b = a + c (a, b, c are in A.P.)


2b – c = a … (ii)


Taking LHS = (a – c)2


= ( 2b – c – c )2 [from eqn. (ii)]


= ( 2b – 2c )2


= 4( b – c )2


= 4( b – c ) ( b – c )


= 4( a – b ) ( b – c ) [b–c = a–b from eqn. (i)]


= RHS


Hence Proved


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


To prove: a2 + c2 + 4ac = 2(ab + bc + ca)


Given: a, b, c are in A.P.


Proof: Since a, b, c are in A.P.


2b = a + c


… (i)


Taking RHS = 2(ab + bc + ca)


Substituting value of b from eqn. (i)






= a2 + c2 + 4ac


= LHS


Hence Proved


(iii) a3 + c3 + 6abc = 8b3


To prove: a3 + c3 + 6abc = 8b3


Given: a, b, c are in A.P.


Formula used: (a+b)3 = a3 + 3ab(a+b) + b3


Proof: Since a, b, c are in A.P.


2b = a + c … (i)


Cubing both side,



8b3 = a3 + 3ac(a+c) + c3


8b3 = a3 + 3ac(2b) + c3 [a+c = 2b from eqn. (i)]


8b3 = a3 + 6abc + c3


On rearranging,


a3 + c3 + 6abc = 8b3


Hence Proved


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