My try : First , I assumed an H.P and then got the answer. Second , tried it by solving but found my method too long in which I have to find the value of a,b,c and d . I want to know how to deal with this problem without assuming H.P.
Pic of question - 
On
a,b,c,d in H.P 1/a,1/b,1/c,1/d ---A.P then
(1/b-1/a) = (1/c-1/b) = (1/d-1/c) =k (a-b)/ab =(c-d)/dc =k--------equ-1 ==(a-b) =kab,(c-d) =kdc
(1/b-1/a) + (1/c-1/b) + (1/d-1/c) =3k (1/d-1/a) = 3k -----------eq---2
(a+b)/(a-b) -(c+d)/(c-d) = (a+b)/kab-(c+d)/kdc ==1/k{(a+b)/ab-(c+d)/dc} 1/k{1/a+1/b-1/c-1/d} 1/k{1/a-1/d+1/b-1/c} 1/k{-3k-k}=-4 ans
Assuming that H.P. means harmonic progression, we have (for $k \neq 0$): $$a=\frac{1}{x}, \quad b=\frac{1}{x+k}, \quad c=\frac{1}{x+2k}, \quad d=\frac{1}{x+3k}.$$ Then $\frac{a+b}{a-b}-\frac{c+d}{c-d}=\frac{\frac{2x+k}{x(x+k)}}{\frac{k}{x(x+k)}}-\frac{\frac{2x+5k}{(x+2k)(x+3k)}}{\frac{k}{(x+2k)(x+3k)}}=\frac{2x+k}{k}-\frac{2x+5k}{k}=\frac{-4k}{k}=-4.$
It is not possible to solve it without the assumption of some progression, the result is easy in the case of harmonic progression.