Let ‘a’ be the first term and ‘d’ the common difference of the given A .P. Then,
a_P = q
⇒ a + (p – 1)d = q ...(i)
a = P
⇒ a + (q – 1 )d = P ...(ii)
Subtracting (ii) from (i), we get
(a – a) + (p – 1)d – (q – 1)d = q – p
⇒ d(p – 1 – q + 1) = q – p
d (p – q) = –(p – q)
d = –1
Putting the value of in (i), we get
a + (p – 1)d = q
⇒ a + (p – 1) (–1) = q
⇒ a – p + 1 = q
⇒ a = p + q – 1
Now, a_n = a + (n – 1)d
= (p + 1) + (n – 1) (1)
= p + q – 1 – n + 1 = p + q – n Proved.