Answer :

Let the first term be ‘a’


& common difference of the given AP is d.


Given: Sm = Sn



2am + md(m – 1) = 2an + nd(n – 1)


2am – 2an + m2d – md – n2d + nd = 0


2a (m – n) + d[(m2 – n2) – (m – n)] = 0


2a (m – n) + d[(m– n)(m + n) – (m – n)] = 0


(m – n) [2a + {(m + n) – 1}d] = 0


2a + (m + n – 1)d = 0 [m – n ≠ 0]…(i)


Now,



[using (i)]


Sm+n = 0


Hence Proved


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