Question

If the $p^{th},q^{th}$ and $r^{th}$ terms of an $A.P$. are $a,b,c$ respectively, then prove that
$\frac{q-r}{bc}+\frac{r-p}{ca}+\frac{p-q}{ab}=0$and $\frac{q-b}{pq}+\frac{b-c}{qr}+\frac{c-a}{rp}=0$

Answer 1

Let 'A' be the first term & 'd' be the common difference

a=A+(p-1)d -(1)

b=A+(q-1)d -(2)

c= A+(r-1)d -(3)

subtracting (2) from (1), (3) from (2), (1) from (3)

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

c(a-b)=c(p-q)d

a(b-c)=a(q-r)d

b(c-a)=b(r-p)d

adding

a(q-r)d+b(r-p)d+c(p-q)d=0

=a(q-r)+b(r-p)+c(p-q)=0

Dividing by abc

$=\frac{(q-r)}{bc}+\frac{(r-p)}{ac}+\frac{(p-q)}{ab}=0$

multiply (4),(5),(6) by r,p,q

r(a-b)=r(p-q)d

p(b-c)=p(q-r)d

q(c-a)=q(r-p)d

q(c-a)=q(r-p)d

adding

= r(a-b)+p(b-c)+q(c-a)=0

dividing by p q r

$=\frac{(a-b)}{pq}+\frac{(b-c)}{qr}+\frac{(c-a)}{pr}=0$