Question

In an AP pth, qth and rth terms are respectively a, b and c. Prove that
p(b - c) + q(c - a) + r(a - b) = 0
(class 10 CBSE SAMPLE PAPER 2017-18 MATHS)

Answer 1

Let A be the first term of an A.P & d be the common difference of the A.P.

Given that,

a = pth term
a= A+(p-1)d……….(1)

b = qth term
b= A+(q-1)d……..(2)

c = rth term

c= A+(r-1)d………….(3)

LHS = (a-b)r +(b-c)p + (c-a)q   (given)

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

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

= -(aq - ar) - (br - bp) - (cp - cq)

= - [a(q - r)+ b(r - p) + c(p - q)]...........(4)

Put the value of a, b,c from eq 1,2,3 in eq4.

= -[(A+(p-1)d)(q - r)+(A+(q-1)d)(r - p)+ (A+(r-1)d)(p - q)]

= - [A(q - r)+(p-1)d(q - r)+ A(r - p) +(q-1)d(r - p) + A(p - q)+(r-1)d(p - q)]

= -[A{(q-r)+ (r-p)+(p - q)}+ d {(p-1)(q - r) +(q-1)(r - p)+(r-1)(p - q)}]

[ Taking A & d common]

= - [A×0+  d {(p-1)(q - r) +(q-1)(r - p)+(r-1)(p - q)}]

=  - [A×0+  d {(pq-pr-q+r)+(qr-pq-r+p)+(rp-qr-p+q)}

=  - [A×0+  d ×0]

= 0×0= 0

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


Hence proved..

HOPE THIS WILL HELP YOU
..

Answer 2

Hey there,

If a,b,c are pth,qth and rth terms of an A.P. ,prove that

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

Solution:

We assume the first term of an A.P.is m, the tolerance = d

a = m + (p-1)*d

b = m + (q-1)*d

c = m + (r-1)*d

p(b-c) = p*(q-r)*d

q(c-a) = q*(r-p)*d

r(a-b) = r*(p-q)*d

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

= p*(q-r)*d + q*(r-p)*d +r*(p-q)*d

= (pq-pr+qr-pq+rp-qr)*d

= 0*d = 0

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

Hope this helps!