Question

The pth, qth and oth terms of an A.P. are a, b and c respectively. Show that a(q – r) + b(r – p) + c(p – 9) = 0.
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Answer 1

Answer:

Complete step-by-step answer:

Let A be the first term D be the common difference of A.P.

The pth term is Tp=a=A+(p−1)D=(A−D)+pD→(1)Tp=a=A+(p−1)D=(A−D)+pD→(1)

The qth term is Tq=b=A+(q−1)D=(A−D)+qD→(2)Tq=b=A+(q−1)D=(A−D)+qD→(2)

The rth term isTr=c=A+(r−1)D=(A−D)+rD→(3)Tr=c=A+(r−1)D=(A−D)+rD→(3)

Here we have got two unknowns A and D which are to be eliminated.

We multiply (1), (2) and (3) by q – r, r – p and p – q respectively and add them together.

⇒a(q−r)+b(r−p)+c(p−q)⇒a(q−r)+b(r−p)+c(p−q)

Substituting a, b and c values from equations (1), (2) and (3)

⇒((A−D)+pD)(q−r)+((A−D)+qD)(r−p)+((A−D)+rD)(p−q)⇒((A−D)+pD)(q−r)+((A−D)+qD)(r−p)+((A−D)+rD)(p−q)

On simplification,

⇒(A−D)[q−r+r−p+p−q]+D[p(q−r)+q(r−p)+r(p−q)]⇒(A−D)[q−r+r−p+p−q]+D[p(q−r)+q(r−p)+r(p−q)]

⇒(A−D)⋅(0)+D[pq−pr+qr−pq+pr−qr]⇒0+D⋅0⇒(A−D)⋅(0)+D[pq−pr+qr−pq+pr−qr]⇒0+D⋅0

= 0

∴a(q−r)+b(r−p)+c(p−q)=0∴a(q−r)+b(r−p)+c(p−q)=0

Hence proved.

Note: We wrote pth, qth and rth terms of A.P. using nth term formula. Then we did algebraic calculation to prove the given condition.