Let the first term be a and 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,
⇒ Sm+n = 0
Hence Proved