Q. 94.4( 5 Votes )

If (b – c)2, (c – a)2, (a – b)2 are in A.P., then show that: are in A.P.

[Hint: Add ab + bc + ca — a2 — b2 — c2 to each term or let = b — c, = c— a, = a — b, then + + = 0]

Answer :

Given: (b – c)2, (c – a)2, (a – b)2 are in A.P


2(c – a)2 = (b – c)2 + (a – b)2 …(i)


To Prove: are in AP


or




2(b – c)(a – b) = (a – c)(c – a)


2[ab – b2 – ca + cb] = ac – a2 – c2 + ac


2ab – 2b2 – 2ac + 2cb = 2ac – a2 – c2


a2 + c2 – 4ac = 2b2 – 2ab – 2cb


Adding both sides, a2 + c2, we get


2(a2 + c2) – 4ac = a2 + b2 – 2ab + c2 + b2– 2cb


2 (a – c)2 = ( b – a)2 + (b – c)2 which is true from (i)


(b – c)2, (c – a)2, (a – b)2 are in A.P


are in AP


Hence Proved


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