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

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Solution

Let A and D be the first term and common difference of the A.P respectively
then ,
$a = \frac{p}{2} [2 A + (p - 1) D] \Rightarrow \frac{a}{p} = \frac{1}{2} [2 A + p D - D]$
$b = \frac{q}{2} [2 A + (q - 1) D] \Rightarrow \frac{b}{q} = \frac{1}{2} [2 A + q D - D]$
$c = \frac{r}{2} [2 A + (r - 1) D] \Rightarrow \frac{c}{r} = \frac{1}{2} [2 A + r D - D]$
So , $\frac{a}{p} (q - r) + \frac{b}{q} (r - p) + \frac{c}{r} (p - q)$
$= \frac{1}{2} [2 A + p D - D] (q - r) + \frac{1}{2} (2 A + q D - D) (r - p) + \frac{1}{2} [2 A + r D - D] (p - q)$
$= \frac{1}{2} ⎡ ⎢ ⎣ 2 A q + p q D - q D - 2 A r - p r D + D r + 2 A r + q D r - D r - 2 A p - q p D + p D + 2 A p + r p D - D p - 2 A q - r q D + q D ⎤ ⎥ ⎦$
$= 0$

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