20. Suppose that $0<a<b<c<d=2 a$ and
$$
(d-a)\left(\frac{a^{2}}{b-a}+\frac{b^{2}}{c-b}+\frac{c^{2}}{d-c}\right)=(a+b+c)^{2}
$$

Find $b c d / a^{3}$.
20. Ans: 4
Set
$$
\begin{aligned}
\mathrm{u} & =(\sqrt{b-a}, \sqrt{c-b}, \sqrt{d-c}) \\
\text { and } \quad \mathrm{v} & =\left(\frac{a}{\sqrt{b-a}}, \frac{b}{\sqrt{c-b}}, \frac{c}{\sqrt{d-c}}\right) .
\end{aligned}
$$

Then $\mathbf{u} \cdot \mathbf{v}=\|\mathbf{u}|\|\mid \mathbf{v}\|$ and hence $\mathbf{u}=\alpha \mathbf{v}$ for some $\alpha \geq 0$. Thus
$$
\frac{b-a}{a}=\frac{c-b}{b}=\frac{d-c}{c} .
$$

Hence $b / a=c / b=d / c$, i.e., $a, b, c, d$ is a geometric progression $a, r a, r^{2} a, r^{3} a=2 a$. Now bed $/ a^{3}=r^{6}=\left(r^{3}\right)^{2}=4$.