# 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]

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

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos  Champ Quiz | Arithmetic Progression34 mins  Champ Quiz | Arithmetic Progression30 mins  Lets Check Your Knowledge in A.P.49 mins  Fundamental Theorem of Arithmetic- 143 mins  NCERT | Fundamental Theorem Of Arithmetic45 mins  Quiz on Arithmetic Progression Quiz32 mins  Arithmetic progression: Previous Year NTSE Questions34 mins  Get to Know About Geometric Progression41 mins  NCERT | Solving Questions on Introduction of A.P42 mins  Arithmetic Progression Tricks and QUIZ37 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 