Question

If \(a, b\) and \(c\) are \(p^{\text {th }}, q^{\text {th }}\) and \(r^{\text {th }}\) terms of an A.P., prove that:

\[ a(q-r)+b(r-p)+c(p-q)=0 \]

Answer 1

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 = \[\frac{(a−b)}{(p−q)}\] ....(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 = \[\frac{b−c}{q−r}\] ....(5)

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

\[\frac{(a−b)}{(p−q)}\] = \[\frac{b−c}{q−r}\]

⇒ (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.