Let first term of A.P. = a
and common difference = d
Then pth term = Tp = a + (p – 1)d = q
⇒ a + (p – 1)d = q … (i)
and qth term = Tp = a + (q – 1)d = p
⇒ a + (q – 1)d = p …. (ii)
(p + qth term =) Tp+q
= a+ {(p + q) -1} …. (iii)
Subtracting equation (ii) from (i),
{a + (p – 1)d} – {a + (q – 1)d} = q – p
⇒ a + (p – 1) d – a – (q – 1)d = q – p
⇒ (p – 1 – q + 1 )d = q – p
⇒ (p – q)d – (p – q)
⇒ d = -1
Putting value of d in equation (i)
a + (p- 1) (- 1) = q
⇒ a – (p – 1) = q
⇒ a = q + p – 1
⇒ a = p + q – 1
Putting the value of a and d in equation (iii),
Tp+q = a + [(P + q) -]d
= (p + q – 1) + [(p + q -1)(- 1)]
= (p + q -1) – (p + q – 1)
= 0
Hence, (p + q)th term is zero.