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) terms is

Let a be the first term and d be the common difference 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