It is given that p,x1,x2,...xn and q,y1,y2,....,yn are in AP and their common difference are a and b. xn=p+na x1=p+a And , yn=q+nb y1=q+b Now, α is the AM of x1,x2,.....xn α=
n
2
(
x1+xn
n
) =
x1+xn
n
And, β is the AM of y1,y2,......,yn β=
y1+yn
2
Therefore, α=
p+a+p+na
2
=
2p+a(n+1)
2
And, β=
q+a+q+na
2
=
2q+b(n+1)
2
Hence,
2α−2p
a
=
2β−2q
b
b(α−p)=a(β−q) Thus, the locus of the point p(α,β) is b(x−p)=a(y−q).