Question

If a,b,c in A.P. (progression) prove that :
(c-a)^2 = 4(a-b)(b-a)

Answer 1

LET a,b,c be a-d,a,a+d put these values for a,b,c in above expression and show L.H.S=R.H.S

Answer 2

мау ве тне 5есоиd bгаскет !и R.H.S = (с-b)

Answer 3

I think we have to use the formula :
b-a=c-b

Answer 4

a-c=0 and a-b=0
a=b=c

Answer 5

A^2+B^2=0 -> A=0 and B=0

Answer 6

(a-c)2+4(a-b)2=0

Answer 7

@mahmit2012
why a-c=0 and a-b=0 in A.P.?

Answer 8

summation of two non negative terms lways is non negative.
So if it is equal to zero, all term should be zero.

Answer 9

but we cant say a-c=0 :/

Answer 10

a b c = 2 4 6 ->

4^2 not equals 4 (-2)(2)

Answer 11

a, a+k, a+2k same problem

Answer 12

it's a question from arithmetic progression..
we can use the formula b-a=c-b to prove ...

Answer 13

Something wrong maybe it is 4(a-b)(a-b) or something...

Answer 14

the previous question was ..
(c-a)^2=4(b^2-ac)
I have done that.

Answer 15

Yes, that one seems to work OK