If `1/a+1/(a-b)+1/c+1/(c-b)= 0` and `a + c-b!=0`, then prove that `a, b, c` are in H.P.

Explanation

We have, `(1)/(a)+(1)/(c)+(1)/(a-b)+(1)/(c-b)=0`
` implies ((1)/(a)+(1)/(c-b))+((1)/(c)+(1)/(a-b))=0`
` implies ((c-b+a))/(a(c-b))+((a-b+c))/(c(a-b))=0`
` implies (a+c-b)[(1)/(a(c-b))+(1)/(c(a-b))]=0`
` implies (a+c-b)[2ac-b(a+c)]=0`
If ` a+c-b ne 0," than "2ac-b(a+c)=0`
or `b=(2ac)/(a+c)`
Therefore,`a,b,c` are in HP and if `2ac-b(a+c) ne 0`, than `a+c-b=0 i.e., b=a+c`.