Let ‘a’ be the 1st term and ‘d’ be the common difference of the A .P.
mam = nan
⇒ m[a + (m – 1) d] = n[a + (n – 1) d]
⇒ m[a + md – d) = n[a + nd – d]
⇒ ma + m2d – md = na + n2d – nd
⇒ ma – na + m2d – n2d – md + nd = 0
⇒ (m – n)a + (m2 – n2)d – d(m – n) =0
⇒ (m – n)a + (m + n) (m – n)d – (m – n)d = 0
⇒ (m – n) [a + (m + n)d – d] = 0
⇒ a + (m + n)d – d = 0 ⇒ a + (m + n – 1)d = 0
⇒ am + n = 0
Thus, the (m + n)th term of the given A .P. is zero.
The first and the last term of an A .P. are 4 and 81 respectively. If the common difference is 7, how many terms are there in the A .P. and what is their sum?