We know that term number n of an A.P is
an=a+(n−1)d
Hence, a=a1+(p−1)d ...(i)
Similarly b=a1+(q−1)d ...(ii)
c=a1+(r−1)d ...(iii)
Subtracting equation (ii) from (i) gives us
a−b=(p−q)d
d=a−bp−q
Substituting in (i) gives us
a=a1+(p−1)(a−b)p−q
a1=a−(p−1)(a−b)p−q
Substituting in (iii), we get
c=a−(p−1)(a−b)p−q+(r−1)(a−b)p−q
a−c=(p−1)(a−b)p−q−(r−1)(a−b)p−q
a−c=a−bp−q(p−r)
(a−c)(p−q)=(a−b)(p−r)
a(p−q)=c(p−q)=a(p−r)−b(p−r)
a(p−q−(p−r))−c(p−q)+b(p−r)=0
a(r−q)+b(p−r)−c(p−q)=0
a(r−q)+b(p−r)+c(q−p)=0