NCERT Solutions
Class 11 Maths
Sequences and Series

Ex.Misc.Q.15
The pth, qth and rth terms of an A.P. are a, b, c respectively.
Show that (q - r)a + (r - p)b + (p - q)c = 0
Let t and d be the first term and the common difference of the A.P. respectively.
The nth term of an A.P. is given by, an = t + (n – 1) d Therefore,
ap = t + (p – 1) d = a ....... (1) aq = t + (q – 1) d = b ...... (2) ar = t + (r – 1) d = c ........ (3)
Subtracting equation (2) from (1), we obtain
(p - 1 – q + 1) d = a – b
(p – q) d = a – b
d = (a – b) ÷ (p - q)
Subtracting equation (3) from (2), we obtain
(p – 1 - r + 1) d = b – c
(q – r) d = b – c
d = (b – c) ÷ (q - r)
Equating both the values of d obtained in (4) and (5), we obtain
(a – b) ÷ (p - q) = (b – c) ÷ (q - r)
(a – b) (q - r) = (b – c) (p - q)
aq – bq – ar + br = bp – bq – cp + cq
bp – cp + cq – aq + ar – br = 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
Thus, the given result is proved.