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Sum of the first p, q and r terms of an A.P. are a, b and c, respectively. Prove that a/p(q - r) + b/q (r- p ) + c/r (p - q) = 0

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Solution:
Let A be the first term and D be the common difference of A.P.

Given that a,b and c be the sum of the first p,q,r terms respectively.

∴  a = Sp

a = ( p/2)[ 2A + ( p -1)D ]

a / p = (1/2)[ 2A + ( p -1)D ] --------------(1)

∴  b = Sq

b = ( q/2)[ 2A + ( q -1)D ]

b / q = (1/2)[ 2A + ( q -1)D ] --------------(2)

∴  c = Sr

c = ( r/2)[ 2A + ( r -1)D ]

c / r = (1/2)[ 2A + ( r -1)D ] --------------(3)

Now a/p(q - r) + b/q (r- p ) + c/r (p - q)

⇒ (1/2)[ 2A + ( p -1)D ](q - r) + (1/2)[ 2A + ( q -1)D ](r- p ) + (1/2)[ 2A + ( r -1)D ](p - q)

⇒ (1/2)( 2A)[ q - r + r- p + p - q)  + (1/2)D [ ( p -1)(q - r) + ( q -1)(r- p ) + ( r -1)(p - q) ]

⇒ (1/2)( 2A) (0) + (1/2)D [ pq - pr - q + r + qr - pq - r + + rp - rq - p + q]

⇒ (0) + (1/2)D (0)

= 0

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