Question

If the sum of p terms of an A.P. is q and the sum of q terms is p, show that the sum of p + q terms is – (p + q). Also, find the sum of first p – q terms (p > q).

Answer 1

Let a be the first term and d the common difference of the given A.P.

∴ S_p = `p/2 [2a + (p - 1)d]` = q

⇒ `2a + (p - 1)d = (2q)/p`  ....(i)

And S_q = `q/2[2a + (q - 1)d]` = p 

⇒ `2a + (q - 1)d = (2p)/q`  ....(ii)

Subtracting equation (ii) from equation (i) we get

(p – q)d = `(2q)/p - (2p)/q`

⇒ (p – q)d = `(2(q^2 - p^2))/(pq)`

⇒ (p – q)d = `(-2)/(pq) (p^2 - q^2)`

⇒ (p – q)d = `(-2)/(pq) (p + q)(p - q)`

⇒ d = `(-2)/(pq) (p + q)`

Substituting the value of d in equation (i) we get

`2a + (p - 1) [(-2(p + q))/(pq)] = (2q)/p`

⇒ 2a = `(2q)/p + (2(p - 1)(p + q))/(pq)`

⇒ a = `q/p + ((p - 1)(p + q))/(pq)`

⇒ a = `(q^2 + p^2 + pq - p - q)/(pq)`

Now S_p+q = `(p + q)/2 [2a + (p + q - 1)d]`

= `(p + q)/2 [(2q^2 + 2p^2 + 2pq - 2p - 2q)/(pq) + ((p + q - 1)[-2(p + q)])/(pq)]`

= `(p + q)/2 [(2q^2 + 2p^2 + 2pq - 2p - 2q - 2p^2 - 2pq + 2p - 2pq - 2q^2 + 2q)/(pq)]`

= `(p + q)/2 [(-2q)/(pq)]`

= `- (p + q)`

Hence proved.