Given: ap = x and aq = y
We know that,
an = a + (n – 1)d
ap = a + (p – 1)d
⇒ x = a + (p – 1)d …(i)
Now,
aq = a + (q – 1)d
⇒ y = a + (q – 1)d …(ii)
From eq. (i) and (ii), we get
x – (p – 1)d = y – (q – 1)d
⇒ x – y = (p – 1)d – (q – 1)d
⇒ x – y = d [p – 1 – q + 1]
⇒ x – y = d[ p – q]
⇒ d = (x - y)/(p - q) …(iii)
Adding, Eq (i) and (ii), we get
x + y = 2a + (p – 1) + (q – 1)d
⇒ x + y = 2a + d[p + q – 1 – 1]
⇒ x + y = 2a + d (p + q – 1) –d
⇒ x + y + d = 2a + (p + q – 1)d …(iv)
We know that,

Hence Proved