Solution. Let $b=a q, c=a q^{2}$ and $d=a q^{3}$. Then

$$
\begin{aligned}
& (a-c)^{2}=\left(a-a q^{2}\right)^{2}=a^{2}-2 a^{2} q^{2}+a^{2} q^{4}=a^{2}\left(1-2 q^{2}+q^{4}\right) \\
& (b-c)^{2}=a^{2}\left(q^{2}-2 q^{3}+q^{4}\right) \\
& (b-d)^{2}=a^{2}\left(q^{2}-2 q^{4}+q^{6}\right) \\
& (a-d)^{2}=a^{2}\left(1-2 q^{3}+q^{6}\right)
\end{aligned}
$$

Substituting the last four equations into the initial expression, we have

$$
\begin{aligned}
(a-c)^{2}+(b-c)^{2}+(b-d)^{2}-(a-d)^{2}= & a^{2}\left(1-2 q^{2}+q^{4}\right)+a^{2}\left(q^{2}-2 q^{3}+q^{4}\right) \\
& +a^{2}\left(q^{2}-2 q^{4}+q^{6}\right)-a^{2}\left(1-2 q^{3}+q^{6}\right) \\
= & a^{2}\left(1-2 q^{2}+q^{4}+q^{2}-2 q^{3}+q^{4}\right. \\
& \left.+q^{2}-2 q^{4}+q^{6}-1+2 q^{3}-q^{6}\right) \\
= & a^{2} \cdot 0=0
\end{aligned}
$$
1A. Let $a, b, c$ and $d$ be four consecutive terms of a geometric progression. Calculate the expression

$$
(a-c)^{2}+(b-c)^{2}+(b-d)^{2}-(a-d)^{2}
$$