Question

please anwer my question !!

Answer 1

it is given that p term=a,q term = b,r term = c
p(b-c)+q(c-a)+r(a-b)
put p=a,q=b and c=r....
a(b-c)+b(c-a)+c(a-b)
solve the bracket ,we get
ab-ac+bc-ba+ca-cb
each one cancelled with one's
0

Answer 2

Hey there !

let a = first term of the AP. and
Letd = common difference of the AP

a= A+(p-1).d........(1)
b= A+(q-1).d........(2)
c= A+(r-1).d..........(3)

subtracting eq. 2 from sq. 1, eq.3 from equipment 2 and equipment 1 from sq 3

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)} .f =0

since d is common difference it should be non zero

Hence

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

Hope it helps you !