Question

If the sum of first m terms of an ap is the same as the sum of its first n terms show that the sum of its first m+n terms is zero

Answer 1

Sum of m terms=Sum of n terms

=> m/2 * (2a + (m-1)d) = n/2 * (2a + (n-1)d)
 Cancelling 2 in the denominator on both sides,

we get

m(2a + (m-1)d) - n(2a+ (n-1)d) = 0
 
 2am + m^2d - md -2an -n^2d +nd =0
 
 2a(m-n) + (m^2 - n^2)d -(m - n)d =0
 
 2a(m-n) + ( (m + n) (m- n) ) d - (m - n )d = 0
 
 Taking (m-n) common
 
 
 2a + ( m + n -1) d = 0 ------------ (1)
 
 
 S m+n = m+n/2( 2a + (m+n -1)d
 
 we know that 2a + (m+n)d is 0 from eqn. 1
 
S m+S n = 0

L.H.S. = R.H.S.

Hence proved