Answer :

Given that a_{p} = 1/q and a_{q} = 1/p

We have to show that S_{pq} = (pq + 1)/2.

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

⇒ 1/q = a + (p – 1) d … (1)

⇒ 1/p = a + (q – 1) d … (2)

Solving (1) and (2),

⇒ (a – a) + (pd – d – qd + d) = 1/q – 1/p

⇒ (p – q) d = 1/q – 1/p

⇒ (p – q) d = (p – q)/ pq

∴ d = 1/pq

Substituting d value in (1),

⇒ a + (p – 1) (1/pq) = 1/q

⇒ a + 1/q – 1/pq = 1/q

⇒ a = 1/q – 1/q + 1/pq

∴ a = 1/pq

We know that the sum of the first n terms of an AP is given by S_{n} = [n (2a + (n – 1) d)] / 2.

Now,

S_{pq} = [(pq) (2(1/pq) + (pq – 1)) (1/pq)] / 2

⇒ S_{pq} =

⇒ S_{pq} = (pq/2) (1/pq + 1)

⇒ S_{pq} = (pq + 1)/ 2

Ans. From the given conditions, we showed that S_{pq} = (pq + 1)/2.

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