Question

If (b-c)^2, (c-a)^2, (a-b)^2 are in AP then 1/(b-c), 1/(c-a), 1/(a-b) are in a) AP b) GP c) HP

Answer 1

Given :   (b-c)², (c-a)², (a-b)² are in AP

To find :  1/(b-c), 1/(c-a), 1/(a-b) are in which Sequence

Solution:

(b-c)², (c-a)², (a-b)² are in AP

Let say

b -c = x

c - a = y

a - b = z

adding all x + y + z = 0

x² , y ² , z²  are in AP

=> y² - x ² =  z² - y²

=> (y + x)(y - x) = (z + y)(z - y)

y + x = - z   *  z + y = - x     (as x + y + z = 0 )

=> (-z)(y - x) = (-x) (z - y)

=> -zy  + zx = -zx + xy

=> 2zx = xy + zy

=> 2zx = y(x + z)

=> 2/y = (x + z)/xz

=> 2/y = 1/z + 1/x

=> 1/x , 1/y , 1/z are in AP

=> 1/(b-c)  , 1/(c -a) , 1/(a - b) are in AP

option A is correct

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