Math, asked by asifkrahim, 2 months ago

The pth, qth and rth term of an AP are a, b and c respectively. Show that a(q-r)+b(r-p)+c(p-q)=0​

Answers

Answered by anant4256kushwaha
1

Answer:

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

T

p

=a=A+(p−1)D=(A−D)+pD (1)

T

q

=b=A+(q−1)D=(A−D)+qD ..(2)

T

r

=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:

a(q−r)+b(r−p)+c(p−q)

=(A−D)[q−r+r−p+p−q]+D[p(q−r)+q(r−p)+r(p−q)]=0.

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