5. Given the arithmetic sequence $\left\{a_{n}\right\}$, the first term and common difference are both positive numbers, 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 $\qquad$.
Answer: 34. Solution: Let the common difference of the sequence $\left\{a_{n}\right\}$ be $d$, then $a_{2}=a_{1}+d, a_{5}=a_{1}+4 d, a_{9}=a_{1}+8 d$. Since $a_{2}, a_{5}, a_{9}$ form a geometric sequence in order, we have $a_{2} a_{9}=a_{5}^{2}$, which means $\left(a_{1}+d\right)\left(a_{1}+8 d\right)=\left(a_{1}+4 d\right)^{2}$. Simplifying the above equation yields: $a_{1} d=8 d^{2}$. Given $d>0$, we have $a_{1}=8 d$. By
$$
\frac{a_{1}+a_{2}+\cdots+a_{k}}{a_{1}}=\frac{a_{1} k+\frac{k(k-1)}{2} d}{a_{1}}=k+\frac{k(k-1)}{16}>100 \text {. }
$$

Solving for $k_{\min }=34$.