vikas askiitian expert
Last Activity: 13 Years ago
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