Suppose x, y, z is a geometric sequence with common ratio r and x≠ y. If x, 2y, 3z is an arithmetic sequence, then r is

1994 AMC 12 Problem 20

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20.

Suppose

x,

y,

z

is a geometric sequence with common ratio

r

and

x\ne y.

x,

2y,

3z

is an arithmetic sequence, then

r

\frac14

\frac13

\frac12

2

4

Write

y=xr

and

z=xr^2

The arithmetic-sequence condition is

4y=x+3z

Because

y=xr

and

z=xr^2,

the arithmetic-sequence condition gives

4xr=x+3xr^2.

Here

x\ne0,

3r^2-4r+1=0,

yielding

r=1

r=\frac{1}{3}.

The condition

x\ne y

excludes

r=1,

leaving

r=\frac{1}{3}.

Thus the correct answer is B.