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

Answer 1

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Answer 2

Given:

In an AP pth , qth and rth terms are respectively a, b and c

To prove:

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

Explanation:

Let A be the first term and D be

the common difference of the given A.P.

Then

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)

On subtracting equation (2) from equation (1),

equation (3) from equation (2)

and

equation (1) from equation (3),

we get

a-b = (p-q)D ----(4)

b-c = (q-r)D -----(5)

and

c-a = (r-p)D -----(6)

Therefore,

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

= (p-q)Dr+(q-r)Dp+(r-p)Dq

/* From (4),(5)and (6) */

= D[ (p-q)r+(q-r)p+(r-p)q]

= D[ pr-qr+pq-pr+qr-pq]

= D×0

= 0

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