Answer :

We use nth term of an A.P. formula

tn = a + (n – 1)d


where n = no. of terms


a = first term


d = common difference


tn = nth terms


Thus mth term = tm = a + (m – 1)d


Given: m × tm = n × tn


m × [ a + (m – 1)d] = n × [a + (n – 1)d]


am + m(m – 1)d = an + n(n – 1)d


am – an + m(m – 1)d – n(n – 1)d = 0


a(m – n) + d[m(m – 1) – n(n – 1)] = 0


a(m – n) + d[ m2 – m – n2 + n] = 0


a(m – n) + d[ (m2 – n2) – m + n] = 0


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


(since, (a – b)(a + b) = a2 – b2)


a(m – n) + d(m – n)[(m + n) –1] = 0


(m – n) [a + d(m + n –1)] = 0


Since, m ≠ n


m – n ≠ 0


a + d(m + n –1) = 0


tm + n = 0


Hence proved


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