If the sum of  m terms of an A.P is the same as the sum of its n  terms, then the sum of its  m+n term is

Let a be the first term and d be the common differences of the given A.P, then Sm=Sn⇒m2[2a+(m−1)d]=n2[2a+(n−1)d]⇒2a(m−n)+{m(m−1)−n(n−1)}d=0⇒2a(m−n)+m2−n2−(m−n)d=0⇒(m−n)[2a+(m+n−1)d]=0⇒2a+(m+n−1)d=0 [∵m−n≠0] Now, Sm+n=m+n2[2a+(m+n−1)d]=m+n2×0=0