Answer :

Sm = n = × [ 2a + (m – 1)d ]


= 2a + (m – 1)d ......................(1)


Sn = m = × [ 2a + (n – 1)d ]


= 2a + (n – 1)d ........................(2)


subtracting both equations, we get :


2( ) = d(m – n)


d = – 2[] ...................(3)


now, Sm + n = × [ 2a + (m + n – 1)d ]


Sm + n × = 2a + (m + n – 1)d .......................(4)


now, (4) – (2), we get :


Sm + n × = d(m)


putting value of d from (3), we get :


Sm + n × = – 2[] × m


Sm + n × = – 2[] × m +


Sm + n × = – 2[] +


Sm + n × = – 2 or


Sm + n = – (m + n)


the correct option is (a).

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