Q. 134.1( 59 Votes )

# If m times

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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