Question

If the relation between `a,b,c` is given as shown below then `a,b,c` are in?

Answer 1

See below.

Explanation:

From

`1/a + 1/(a - b) + 1/c + 1/(c - b)=0` we conclude

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

but

`(a - b + c) ne 0 rArr a (b - 2 c) + b c =0 rArr ab-ac = ac-bc`

and after dividing both sides by `abc`

`1/a-1/b = 1/b-1/c rArr` HP

Answer 2

Given

`1/a+1/(a-b)+1/c+1/(c-b)=0`

`=>1/a+1/(c-b)+1/c+1/(a-b)=0`

`=>(a+c-b)/(a(c-b))+(a+c-b)/(c(a-b))=0`

`=>(a+c-b)[1/(a(c-b))+1/(c(a-b))]=0`

As `(a+c-b)!=0`

`1/(a(c-b))+1/(c(a-b))=0`

`=>(a(c-b))=-c(a-b))`

`=>ac-ab=-ac+bc`

`=>2ac=ab+bc`

`=>(2ac)/(abc)=(ab)/(abc)+(bc)/(abc)`

`=>2/b=1/c+1/a`

This means `1/a,1/b,1/c "are in AP"`

So `a,b,c` are in HP, option ( C)