If a, b and c are p^th, q^th and r^th terms of an A.P., prove that

a(q - r) + b(r − p) + c(p − q) = 0

Let t and d be the first term and common difference of the A.P respectively.

The nth term of an A.P is given by, a_n = t + (n - 1) d

pth term = t + (p - 1)d = a ….(1)

qth term = t + (q - 1)d = b ….(2)

rth term = t + (r - 1)d = c ….(3)

Subtracting (2) from (1), we obtain :

⇒ t + (p - 1)d - [t + (q - 1)d] = a - b

⇒ t + (p - 1)d - t - (q - 1)d = a - b

⇒ (p - 1 - q + 1)d = a - b

⇒ (p - q)d = a - b

⇒ d = $\dfrac{(a - b)}{(p - q)}$ ….(4)

Subtracting (3) from (2), we obtain :

⇒ t + (q - 1)d - [t + (r - 1)d] = b - c

⇒ t + (q - 1)d - t - (r - 1) d = b - c

⇒ (q - 1 - r + 1)d = b - c

⇒ (q - r)d = b - c

⇒ d = $\dfrac{b - c}{q - r}$ ….(5)

From (4) and (5), we get:

$\dfrac{(a - b)}{(p - q)} = \dfrac{b - c}{q - r}$

⇒ (a - b)(q - r ) = (b - c)( p - q)

⇒ aq - ar - bq + br = bp - bq - cp + cq

⇒ bp - cp + cq - aq + ar - br - bq + bq = 0

⇒ (-aq + ar) + (bp - br) + (-cp + cq) = 0

⇒ -a(q - r) - b(r - p) - c(p - q) = 0

⇒ a(q - r) + b(r - p) + c(p - q) = 0

Hence, proved that a(q - r) + b(r − p) + c(p − q) = 0.