If S_m = n and S_n = m then S_{m + n} = ……

⇒ S_m = n = × [ 2a + (m – 1)d ]

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

⇒ S_n = m = × [ 2a + (n – 1)d ]

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

subtracting both equations, we get :

⇒ 2( – ) = d(m – n)

⇒ d = – 2[] ...................(3)

now, S_{m + n} = × [ 2a + (m + n – 1)d ]

⇒ S_{m + n} × = 2a + (m + n – 1)d .......................(4)

now, (4) – (2), we get :

⇒ S_{m + n} × – = d(m)

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

⇒ S_{m + n} × – = – 2[] × m

⇒ S_{m + n} × = – 2[] × m +

⇒ S_{m + n} × = – 2[] +

⇒ S_{m + n} × = – 2 or

⇒ S_{m + n} = – (m + n)

∴ the correct option is (a).