Given, mtm = ntn
m[a + (m – 1) d] = n [a + (n – 1) d]
[we know that tn = a + (n – 1)d]
m[a + md – d] = n[a + nd – d]
ma + m2d – md = na + n2d – nd
ma – na + m2d – n2d = md – nd
a (m – n) + d (m2 – n2) = d(m – n)
a (m – n) + d(m + n)(m – n) = d(m – n) ÷ by (m – n) on both sides,
a + d (m + n) = d
a + d(m + n) – d = 0
a + d(m + n – 1) = 0 … (1)
To prove, tm + n = 0
tm + n = a + (m + n – 1)d
tm + n = 0 (from(1))
Hence it is proved.