Question

10. If pth, qth and rth terms of an H.P. are a, b and e respectively, then prove that : bc(q – r) + ca(r - p) + ab(p - q) = 0

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Answer 1

Answer:

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

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Step-by-step explanation:

Answer 2

Answer:

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

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