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The pth , qth and rth terms of an A.P are a, b,c respectively . Show that (q – r)a + (r – p)b + (p - q) c = 0

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Let A, A + D, A + 2D,…….. are in A.P.

Given pth term = A + (p – 1 )D = a ……………. (1)

qth term = A + (q – 1)D = b …………………. (2)

and rth term = A + (r - 1)D = C …………………(3)

X(q - r) ⇒ a(q – r) = (q – r) [A + (p – 1)D] 

X(r – p) ⇒ b(r - p) = (r - p)[A + (q – 1)D] x (p - q)

⇒ c(p - q) = (p - q)[A + (r – 1)D]

By adding we get

⇒ a(q - r) + b(r – p) + c(p – q)

= A[(q - r) + (r - p) + (p - q) + D] [(p - 1)(q – r) + (q – 1 )(r – p) + (r - 1)(p – q)] 

= A(0) + D(0) = 0

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