let common differnece of this series is d & first term is a then
 
Sm = n = m/2 [ 2a + (m-1)d ] 
       2n/m = 2a + (m-1)d                                ......................1
 
Sn = m = n/2 [ 2a + (n-1)d ]                 
      2m/n = 2a + (n-1)d                               ........................2
 
subtracting both equations ,
2(n/m - m/n) = d(m-n)
 d = -2[n+m/nm]                 ...................3
 
now , Sm+n = (m+n)/2 [ 2a + (m+n-1)d ]
      Sm+n (2/m+n) = 2a + (m+n-1)d               .......................4
eq 4 - eq 2
 Sm+n (2/m+n) - 2m/n =  d(m)   
putting value of d from eq 3 we get
Sm+n (2/m+n) = -2            or
Sm+n = -(m+n)              
this is the required result
 
						

							Sm=n
m/2(2a+(m-1)d)=n
2a+md-d=2n/m......eqn 1.

Sn=m
Similarly, 2a+nd-d=2m/n....eqn2.

Subtracting 2 from 1, we get-
d=-2(m+n)/mn

Sub. this value of d in eqn 1.
2a+md-d=2n/m
[2a+m*-2(m+n)+2(m+n)]/mn=2n/m
Making 2a the subject of the eqn, we get-
2a=[2n(sq)+2m(sq)+2mn-2m-2n]/mn.......(sq) stands for square, ex:2n(sq)=2n*n....pls note!

S(m+n)=(m+n)/2*[2a+(m+n-1)d]
Now substitute the value of 2a and d we got earlier in the above eqn-
You will get-
(m+n)/2*[2n*n+2m*m+2mn-2m-2n-2m*m-2n*n-4mn+2m+2n]/mn.....
The final eqn you get on simplification is-
(m+n)/2*(-2mn)/m

The Answer is -m-n....
Be very carefull while simplification as its a lenghty problem...