Question

The sum of nth term of an ap is m and it's mth term is n then show that the sum of m+n terms is 0

Answer 1

Sum of nth term = n/ 2 [ 2a + ( n - 1 ) d] = m

Sum of mth terms = m/2 [ 2a + ( m - 1 ) d] = n

S( m + n ) = ( m + n ) /2 [ 2a + ( m + n - 1) d]

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

2an + n^2d - nd = 2am + m^2 d - md

2an - 2am = m^2 d - n^2 d +nd - md

2a ( n - m ) = d( m^2 - n^2 + n - m )

2a ( n - m ) = d [ ( m - n ) ( m + n ) + ( n - m )]

-2a ( m - n ) = d [ ( m - n ) ( m + n ) - ( m n )]

-2a ( m - n ) = d ( m - n) [ ( m +n - 1 )]

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

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

Putting value of d,

S ( m + n ) = ( m + n) /2 [ 2a + ( m + n - 1 ) ( - 2a) / ( m + n - 1 )]

S ( m +n) = ( m +n ) /2 [ 2a - 2a ]

S ( m + n ) = ( m +n ) /2 (0)

S(m + n) = 0

Hence, Proved.