Question

The pth, qth and rth terms of an A.P are a,b and c Respectively. show that a(q-r)+b(r-p)+c(p-q)=0.​

Answer 1

I have attached the solution

HOPE THIS HELPS

Answer 2

Let a = first term of the AP.

and

Let d = common difference of the AP

Now

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

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

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

Subtracting 2nd from 1st , 3rd from 2nd 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)

Let a = first term of the AP.

and

Let d = common difference of the AP

Now

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

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

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

Subtracting 2nd from 1st , 3rd from 2nd 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(6)