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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).
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Solution
Let a be the first term and d the common difference of the given A.P.
∴ Sp = `p/2 [2a + (p - 1)d]` = q
⇒ `2a + (p - 1)d = (2q)/p` ....(i)
And Sq = `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 Sp+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.
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