Question

the p and q term of series are a,b,c respectively proof that ( a-r )a+(r-p)b+(p-q)c=0​

Answer 1

Step-by-step explanation:

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.

Answer 2

Step-by-step explanation:

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.