Problem:
Suppose x,y,z is a geometric sequence with common ratio r and xξ =y. If x,2y,3z is an arithmetic sequence, then r is
Answer Choices:
A. 41β
B. 31β
C. 21β
D. 2
E. 4
Solution:
Since y=xr,z=xr2 and 3zβ2y=2yβx, by substitution
3xr2β2xr=2xrβx or 3r2β4r+1=0
Thus (3rβ1)(rβ1)=0. Since xξ =y, it follows that rξ =1. Thus r=1/3.