On a geometric sequence

Let $\alpha, \beta, \gamma , \delta$ be four consecutive terms of a geometric sequence. Prove that

$\left ( \beta -\gamma \right )^2+\left ( \gamma -\alpha \right )^2+\left ( \delta -\beta \right )^2 = \left ( \alpha -\delta \right )^2$

Solution

We have successively:

$\begin{align*} \left ( \beta -\gamma \right )^2+\left ( \gamma -\alpha \right )^2+\left ( \delta -\beta \right )^2 &= \beta^2 -2\beta \gamma + \gamma^2 + \gamma^2 - 2 \gamma \alpha + \alpha^2 + \delta^2 - 2\delta \beta + \beta^2 \\ &= \alpha \gamma - 2\beta \gamma + \gamma^2 +\gamma^2 - 2\gamma \alpha + \alpha^2 + \delta^2 - 2 \delta \beta+ \alpha \gamma \\ &= -2\beta \gamma + 2 \delta \beta + \alpha^2+ \delta^2 - 2\delta \beta \\ &= \alpha^2 -2 \beta\gamma + \delta^2 \\ &= \alpha^2 - 2\alpha \delta + \delta^2 \\ &= \left ( \alpha - \delta \right )^2 \end{align*}$