If the p^(th), q^(th) and r^(th) terms of an arithmetic progression are P, Q, R respectively, then P(q-r) + Q(r-p) + R(p-q) is equal to?

Nov 29, 2016

$P \left(q - r\right) + Q \left(r - p\right) + R \left(p - q\right) = 0$

Explanation:

The ${n}^{\text{th}}$ term ${a}_{n}$ of an arithmetic progression with common difference $d$ and initial term ${a}_{1}$ is given by

${a}_{n} = {a}_{1} + \left(n - 1\right) d$

thus, we can write

$P = {a}_{1} + \left(p - 1\right) d$
$Q = {a}_{1} + \left(q - 1\right) d$
$R = {a}_{1} + \left(r - 1\right) d$

If we subtract $R$ from $Q$, we find

$Q - R = {a}_{1} + \left(q - 1\right) d - \left({a}_{1} + \left(r - 1\right) d\right) = \left(q - r\right) d$

Thus $q - r = \frac{Q - R}{d}$

Similarly, we have $r - p = \frac{R - P}{d}$ and $p - q = \frac{P - Q}{d}$

Substituting these into the given expression, we get

$P \left(q - r\right) + Q \left(r - p\right) + R \left(p - q\right)$

$= \frac{P \left(Q - R\right)}{d} + \frac{Q \left(R - P\right)}{d} + \frac{R \left(P - Q\right)}{d}$

$= \frac{P Q - P R + Q R - P Q + P R - Q R}{d}$

$= \frac{P Q - P Q + P R - P R + Q R - Q R}{d}$

$= \frac{0}{d}$

$= 0$