Sm=n , Sn = m then S(m+n) = ?

let common differnece of this series is d & first term is a then

 

S_m = n = m/2 [ 2a + (m-1)d ] 

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

 

S_n = 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 , S_m+n = (m+n)/2 [ 2a + (m+n-1)d ]

      S_m+n (2/m+n) = 2a + (m+n-1)d               .......................4

eq 4 - eq 2

 S_m+n (2/m+n) - 2m/n =  d(m)   

putting value of d from eq 3 we get

S_m+n (2/m+n) = -2            or

S_m+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...