9. Given that $a, b, c, d$ are even numbers, and $0<a<b<c<d, d-a=90, a, b, c$ form an arithmetic sequence, $b, c, d$ form a geometric sequence. Then $a+b+c+d=$ ( ).
(A) 384
(B) 324
(C) 284
(D) 194
9.D.

Let $a, b, c, d$ be $b-m, b, b+m, \frac{(b+m)^{2}}{b}$, respectively.
Also, $\frac{(b+m)^{2}}{b}-(b-m)=90$, which means
$$
b=\frac{m^{2}}{3(30-m)} \text {. }
$$

Since $a, b, c, d$ are even numbers, and $0<a<b<c<d$, we know that $m$ is a multiple of 6, and $m<30$.
Let $m=6k$, substituting into equation (1) yields
$$
b=\frac{2 k^{2}}{5-k}(k=1,2,3,4) \text {. }
$$

Substituting and checking, we find $k=4, b=32$.
Thus, $m=24, b=32, a, b, c, d$ are $8, 32, 56, 98$ respectively.
Therefore, $a+b+c+d=194$.