The sum of the first p terms of an A.P. is q and the sum of the first

1.	The sum of the first p terms of an A.P. is q and the sum of the first q terms is p. Find the sum of the first (p + q) terms.(a) pq (b) p – q (c) – (p + q) (d) 0
Answer» (c) - (p + q) Since S_n = \(\frac{n}{2}\)[2a + (n - 1) d] for an A.P. whose first term = a, common difference = d, number of terms = n. ∴ S_p = q = \(\frac{p}{2}\) (2a + (p – 1)d) ⇒ 2q = 2ap + p (p – 1)d ...(i) S_q = p = \(\frac{q}{2}\) (2a + (q – 1) d) ⇒ 2p = 2aq + q(q – 1)d ...(ii) Subtracting eqn (ii) from eqn (i), we get 2(q – p) = 2a (p – q) + (p² – q²)d – (p – q) d ⇒ – 2(p – q) = 2a (p – q) + (p – q) (p + q) d – (p – q) d – 2 = 2a + [(p + q) – 1]d ...(iii) Now, S_{p + q}= \(\frac{p+q}{2}\) [2a + (p + q - 1) d ] = \(\frac{p+q}{2}\) x - 2 (From (iii)) = - (p + q).

The sum of the first p terms of an A.P. is q and the sum of the first q terms is p. Find the sum of the first (p + q) terms.(a) pq (b) p – q (c) – (p + q) (d) 0

Answer»

(c) - (p + q)

Since S_n = \(\frac{n}{2}\)[2a + (n - 1) d] for an A.P. whose first term

= a, common difference = d, number of terms = n.

∴ S_p = q = \(\frac{p}{2}\) (2a + (p – 1)d)

⇒ 2q = 2ap + p (p – 1)d ...(i)

S_q = p = \(\frac{q}{2}\) (2a + (q – 1) d)

⇒ 2p = 2aq + q(q – 1)d ...(ii)

Subtracting eqn (ii) from eqn (i), we get

2(q – p) = 2a (p – q) + (p² – q²)d – (p – q) d

⇒ – 2(p – q) = 2a (p – q) + (p – q) (p + q) d – (p – q) d

– 2 = 2a + [(p + q) – 1]d ...(iii)

Now, S_{p + q}= \(\frac{p+q}{2}\) [2a + (p + q - 1) d ]

= \(\frac{p+q}{2}\) x - 2 (From (iii))

= - (p + q).