$10 \cdot 37$ Three non-zero numbers $a, b, c$ form an arithmetic sequence. When $a$ is increased by 1 or $c$ is increased by 2, these three numbers form a geometric sequence. Then $b$ equals
(A) 16.
(B) 14.
(C) 12.
(D) 10.
(E) 8.
(14th American High School Mathematics Examination, 1963)
[Solution] Given $2 b=a+c$, and $b^{2}=c(a+1)$, also $b^{2}=a(c+2)$. $\therefore c=2 a$, then $a=\frac{2}{3} b$.
$\therefore b^{2}=\frac{8}{9} b^{2}+\frac{4}{3} b$, solving this gives $b=12$. Therefore, the answer is $(C)$.