if the pth qth and rth terms of an ap be a b and

Question

if the pth qth and rth terms of an ap be a b and c respectively then show that a(q-r)+b(r-p)+c(p-q)=0

Vikas TU · Accepted Answer

Dear student

Let a = first term of the AP.
and
Let d = common difference of the AP

Now
a = A+(p-1).d.......(i)
b = A+(q-1).d.......(ii)
c = A+(r-1).d........(iii)

Subtracting 2nd from 1st , 3rd from 2nd and 1st from 3rd we get

a-b = (p-q).d......(iv)
b-c = (q-r).d........(v)
c-a = (r-p).d.......(vi)

multiply iv,v,vi by c,a,b respectively we have

c.(a-b) = c.(p-q).d......(vii)
a.(b-c) = a.(q-r).d........(viii)
b.(c-a) = b.(r-p).d.......(ix)

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

if the pth qth and rth terms of an ap be a b and c respectively then show that a(q-r)+b(r-p)+c(p-q)=0