$10 \cdot 40$ If $x, y, z$ are terms of a geometric sequence with a common ratio $r$, and $x \neq y$; also, $x, 2y, 3z$ form an arithmetic sequence. Then $r=$
(A) $\frac{1}{4}$.
(B) $\frac{1}{3}$.
(C) $\frac{1}{2}$.
(D) 2 .
(E) 4 .
(45th American High School Mathematics Examination, 1994)
[Solution] Let $y=r x, z=r^{2} x$, and $r \neq 1$. Also, $x, 2 y, 3 z$ form an arithmetic sequence, so $2 y=\frac{1}{2}(x+3 z)$, which means $r x=\frac{1}{2}\left(x+3 r^{2} x\right)$ or $3 r^{2}-4 r+1=0$. Solving this, we get $r_{1}=1, r_{2}=\frac{1}{3}$. Since $r \neq 1$, we have $r=\frac{1}{3}$. Therefore, the answer is (B).