5. Given the sequence $\left\{a_{n}\right\}$ is an arithmetic sequence, the first term and common difference are both positive, and $a_{2}, a_{5}, a_{9}$ form a geometric sequence in order, then the smallest positive integer $k$ such that $a_{1}+a_{2}+\cdots+a_{k}>100 a_{1}$ is $\qquad$
Let the common difference of the sequence $\left\{a_{n}\right\}$ be $d, a_{5}^{2}=a_{2} a_{9} \Rightarrow\left(a_{1}+4 d\right)^{2}=\left(a_{1}+d\right)\left(a_{1}+8 d\right)$
$$
\begin{array}{l}
\Rightarrow a_{1}=8 d \Rightarrow a_{n}=(n+7) d, n \in \mathbf{N}^{*} . \\
\text { Thus } a_{1}+a_{2}+\cdots+a_{k}=\frac{k(k+15) d}{2}>100 a_{1}=800 d \Rightarrow k(k+15)>1600 \\
\Rightarrow k>\frac{\sqrt{15^{2}+6400}-15}{2}=\frac{5 \sqrt{256+9}-15}{2}=\frac{5 \cdot 16\left(1+\frac{9}{256}\right)^{\frac{1}{2}}-15}{2} \\
>\frac{5 \cdot 16\left(1+\frac{9}{512}\right)-15}{2}=\frac{80+\frac{45}{32}-15}{2}>33 \Rightarrow k \geqslant 34 .
\end{array}
$$

Therefore, the smallest positive integer $k$ such that $a_{1}+a_{2}+\cdots+a_{k}>100 a_{1}$ is 34.