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

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Solution

We know that $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
Sum of first p terms $= a \Rightarrow S_{p} = a$
$\Rightarrow \frac{P}{2} [2 a_{1} + (P - 1) d] = a . . . (1)$
where $a_{1}$ is the first term
Sum of first q terms $= b =\Rightarrow S_{q} = b$
$= \frac{q}{2} [2 a_{1} + (q - 1) d] = b . . . . (2)$
Sum of first r terms $= c \Rightarrow S_{r} = c$
$\Rightarrow \frac{r}{2} [2 a_{1} + (r - 1) d] = c . . . (3)$
$(1) \Rightarrow \frac{a}{p} = a_{1} + (\frac{P - 1}{2}) d - (4) (2) \Rightarrow \frac{b}{q} = a_{1} (\frac{a - 1}{2}) . . . (5)$
$(3) \Rightarrow \frac{c}{r} = a_{1} + (\frac{r - 1}{2}) d . . . (6)$
From (4),(5) & (6) we have
LHS $= \frac{a}{p} (q - r) + \frac{b}{q} (r - p) + \frac{c}{r} (p - q)$
$= (a_{1} (\frac{p - 1}{2}) d) (q - r) + (a_{1} + (\frac{q - 1}{2})) (r - 1) + (a_{1} + (\frac{r - 1}{2} d)) (p - q)$
$= a^{1} (q - r + r - p + p - q) + \frac{d}{2} [(p - 1) (q - r) + (q - 1) (r - p) + (r - 1) (p - q)]$
$= a_{1} (0) + \frac{d}{2} [p q - p r - q + r + q r - q r - q p - r + p + r p - r q - p + q]$
$= a_{1} (0) + \frac{d}{2} (0) = 0$
Hence proved

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