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Question

$p^{t h}$ , $q^{t h}$ and $r^{t h}$ terms of an arithmetic progression are $a, b, c$ respectively then prove that $a (q - r) + b (r - p) + c (p - q) = 0$

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Solution

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Let the first term and common difference of the A.P be A and D respectively
then
$a = A + (p - 1) D$
$b = A + (q - 1) D$
$c = A + (r - 1) D$
Now,
$a (q - r) + b (r - b) + c (p - q)$
$= [A + (p - 1) D] (q - r) + [A + (q - 1) D (r - p)] + [A + (r - 1) D] (p - q)$
$= A q - A r + p d q - p s r - D q + D r + A r - A p + q d r - q d p - d r + d p$

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