Question

28. If (b - c)^2, (c-a)^2, (a - b)^2 are in A.P. then (b - c), (c-a), (a - b) are in
(a) A.P.
(b) G.P.
(c) H.P.
(d) None​

Answer 1

Answer:

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

⇒2b=a+c……. (1)

And a

2

,b

2

,c

2

are in H. P.

b

2

1

−

d

2

1

=

c

2

1

−

b

2

1

b

2

a

2

(a–b)(a+b)

=

b

2

c

2

(b–c)(b+c)

ac

2

+bc

2

=a

2

b+a

2

c[∵a–b=b–c]

ac(c–a)+b(c–a)(c+a)=0

(c–a)(ab+bc+ca)=0

either c–a=0orab+bc+ca=0

either c=a or (a+c)b+ca=0 and then form (i) 2b

2

+ca=0

Either a=b=c or b

2

=a

2

−c

i.e. a,b,

2

−c

are in G. P. Hence Proved

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