In $\triangle A B C$, the three sides $a, b, c$ form an arithmetic sequence. Find the value of $5 \cos A-4 \cos A \cos C+5 \cos C$.
Three, from the condition, $2 b=a+c$,
$$
\begin{array}{l}
\therefore 2 \sin B=\sin A+\sin C . \\
\text { Also } A+B+C=\pi, \therefore 2 \sin (A+C)=\sin A+\sin C . \\
\therefore 4 \sin \frac{A+C}{2} \cos \frac{A+C}{2}=2 \sin \frac{A+C}{2}=\cos \frac{A-C}{2} . \\
\text { Since } \sin \frac{A+C}{2} \neq 0, \therefore 2 \cos \frac{A+C}{2}=\cos \frac{A-C}{2} .
\end{array}
$$

Thus, $5 \cos A-4 \cos A \cos C+5 \cos C$
$$
\begin{array}{l}
=5(\cos A+\cos C)-4 \cos A \cos C \\
=10 \cos \frac{A+C}{2} \cos \frac{A-C}{2}-2[\cos (A+C)+\cos (A-C)] \\
=10 \cos \frac{A+C}{2} \cos \frac{A-C}{2}-2\left[2 \cos ^{2} \frac{A+C}{2}+8 \cos ^{2} \frac{A+C}{2}-2\right] \\
=20 \cos ^{2} \frac{A+C}{2}-20 \cos ^{2} \frac{A+C}{2}+4=4 .
\end{array}
$$