If the p^th term of an A. P. is q and q^th term is p, prove that its n^th term is (p + q – n).

Given that a_p = q and a_q = p

We have to show that a_n = (p + q – n).

We know that in an AP with first term a and common difference d, the nth term (or the general term) is given by a_n = a + (n – 1) d.

∴ a_p = a + (p – 1) d and a_q = a + (q – 1) d

⇒ q = a + (p – 1) d … (1)

⇒ p = a + (q – 1) d … (2)

Solving (1) and (2),

⇒ (a – a) + (pd – d – qd + d) = q – p

⇒ (p – q) d = q – p

⇒ (p – q) d = -(p – q)

∴ d = -1

Substituting d value in (1),

⇒ a + (p – 1) (-1) = q

⇒ a – p + 1 = q

⇒ a = q + p – 1

∴ a = p + q - 1

Now, a_n = a + (n – 1) d

Substituting a and d values,

⇒ a_n = (p + q – 1) + (n – 1) (-1)

⇒ a_n = p + q – 1 – n + 1

∴ a_n = p + q - n

Ans. Hence from the given conditions we showed that a_n = (p + q – n).