14. If $a, b, c$ form an arithmetic sequence, then the midpoint of the line segment cut by the line $a x + b y + c = 0$ on the ellipse $\frac{x^{2}}{2} + \frac{y^{2}}{8} = 1$ has the trajectory equation $\qquad$.
$14.2\left(x-\frac{1}{2}\right)^{2}+\frac{(y+1)^{2}}{2}=1$.
From $a-2 b+c=0$, we know that the line $a x+b y+c=0$ passes through the fixed point $P(1,-2)$. Since point $P$ lies on the ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{8}=1$, $P$ is one endpoint of the intercepted segment. Let the other endpoint be $Q\left(x_{1}, y_{1}\right)$, and the midpoint of segment $P Q$ be $M\left(x_{0}, y_{0}\right)$, then
$$
\left\{\begin{array} { l } 
{ x _ { 0 } = \frac { x _ { 1 } + 1 } { 2 } , } \\
{ y _ { 0 } = \frac { y _ { 1 } - 2 } { 2 } , }
\end{array} \text { i.e., } \left\{\begin{array}{l}
x_{1}=2 x_{0}-1 \\
y_{1}=2 y_{0}+2 .
\end{array}\right.\right.
$$

Since point $Q\left(x_{1}, y_{1}\right)$ lies on the ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{8}=1$, we have
$$
\frac{\left(2 x_{0}-1\right)^{2}}{2}+\frac{\left(2 y_{0}+2\right)^{2}}{8}=1 \text {. }
$$

Therefore, the equation of the trajectory of the midpoint $M$ is $2\left(x-\frac{1}{2}\right)^{2}+\frac{(y+1)^{2}}{2}=1$.