Question

Question 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

Class X1 - Maths -Sequences and Series Page 199

Answer 1

Let A and d be ist term and common difference of the series AP

Then

a = A+(p-1).d.......(1)

b = A+(q-1).d.......(2)

c = A+(r-1).d........(3)

subtracting 2 from 1, 3 from 2 and 1st from 3rd we get

a-b = (p-q).d......(4)

b-c = (q-r).d........(5)

c-a = (r-p).d.......(6)

multiply 4,5,6 by c,a,b respectively we have

c.(a-b) = c.(p-q).d......(4)

a.(b-c) = a.(q-r).d........(5)

b.(c-a) = b.(r-p).d.......(6)

a(q-r).d+b(r-p).d+c(p-q).d = 0

(a(q-r)+b(r-p)+c(p-q)).d = 0

Now since d is common difference it should be non zero hence

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

Answer 2

Let A is the first term and D is the common difference of Given AP.
now,
Pth term = A + (P - 1)D = a -------------(1)
qth term = A + (q - 1)D = b -------------(2)
rth term = A + ( r - 1)D = c -------------(3)

Now,
LHS = (q - r)a + (r - p)b + (p - q)c
from equations (1) , (2) and (3)
= (q - r)[A + (p -1)D ] + (r - p)[A + (q - 1)D ] + (p - q)[A + (n - 1)D ]
= A { (q - r) + ( r - p) + ( p - q) } + D{ (q - r)(p -1) + ( r - p)(q - 1) + ( p - q)( r - 1)}
= A × 0 + D [ qp - q - rp + r + rq - r - pq + p + pr - p - qr + q ]
= 0 + D × 0
= 0 = RHS