Let OP=x+yi and p 1 point coordinates (x 1, y 1).

x 1 =( 1+0)/2 = 1/2，y 1 =(0+ 1)/2 = 1/2

OA= 1 OB=i

From the meaning of the question, get

x+yi=an+bni

x=an

y=bn

x 1 = a 1y 1 = b 1

a 1 = 1/2 b 1 = 1/2

Let arithmetic progression {an} tolerance be d d≠0, then the general formula is an =1/2+(n-1) d.

n=(x- 1/2)/d + 1

Let the common ratio {bn} of geometric series be q, then the general formula is BN = (1/2) q (n- 1).

q^(n- 1)=2y

n=[lg(2y)/lgq]+ 1

(x- 1/2)d=[lg(2y)/lgq]

(x- 1/2)dlgq=lg(2y)

y=q^(x- 1/2)d/2

There is a unique solution to any non-zero d and q, only q= 1, in which case y=d/2, and for a given d, y is always d/2.

That is, there is a unique geometric series (actually a constant sequence) bn= 1/2, which satisfies the problem.