Suppose the sum of the first m terms of an arithmetic progression is n and the sum of its first n terms is m where m ≠ n . Then, the sum of the first ( m + n ) terms of the arithmetic progression is

Step-by-step solution

Given, S m = n and S n = m S m = m 2 [ 2 a + ( m - 1 ) d ] = n ...( i ) S n = n 2 ( 2 a + ( n - 1 ) d ) = m ...( ii ) On subtracting Eq. ( ii ) from Eq. ( i ), we get ( m - n ) a + ( m - n ) ( m + n - 1 ) d 2 = - ( m - n ) ⇒    2 a + ( m + n - 1 ) d = - 2    [ m ≠ n ] ∴    S m + n = m + n 2 ( 2 a + ( m + n - 1 ) d ) = m + n 2 ( - 2 ) = - ( m + n )