Answer :

We use n^{th} term of an A.P. formula

t_{n} = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_{n} = n^{th} terms

Thus m^{th} term = t_{m} = a + (m – 1)d

Given: m × t_{m} = n × t_{n}

⇒ 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[ m^{2} – m – n^{2} + n] = 0

⇒ a(m – n) + d[ (m^{2} – n^{2}) – m + n] = 0

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

(since, (a – b)(a + b) = a^{2} – b^{2})

⇒ 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

⇒ t_{m + n} = 0

Hence proved

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