7. Given a sequence of positive numbers $a_{1}, a_{2}, \ldots, a_{10}$, satisfying the relation $a_{n}\left(a_{n-1}+a_{n+1}\right)=2 a_{n-1} a_{n+1}\left(a_{n}+1\right)$ for $n=2,3, \ldots, 9$. Find $a_{5}$ if it is known that $a_{1}=1$ and $a_{10}=0.01$
Solution: Divide both sides by $a_{n-1} a_{n} a_{n+1} \neq 0$ (since the numbers are positive) and make the substitution $\frac{1}{a_{n}}=b_{n}$, we get the relation $b_{n-1}+b_{n+1}=2 b_{n}+2 \Leftrightarrow b_{n+1}-b_{n}=b_{n}-b_{n-1}+2$. It is clear that the distances between adjacent terms of the sequence $b_{n}$ increase by 2 each time. If we denote $b_{2}-b_{1}=d$, then these distances will be equal to $d, d+2, \ldots, d+16$. And their sum $\left(b_{2}-b_{1}\right)+$ $\cdots+\left(b_{10}-b_{9}\right)=b_{10}-b_{1}=100-1=99$. We get the equation $d+\cdots+(d+16)=9 d+72=$ 99, from which $d=3$. We get $b_{2}=4, b_{3}=9, b_{4}=16, b_{5}=25 \Rightarrow a_{5}=\frac{1}{25}$

Answer: $0.04$.