6. (2001 National High School Competition Question) Let $\left\{a_{n}\right\}$ be an arithmetic sequence, and $\left\{b_{n}\right\}$ be a geometric sequence, and $b_{1}=a_{1}^{2}, b_{2}=a_{2}^{2}$, $b_{3}=a_{3}^{2}\left(a_{1}<a_{2}\right)$, and $\lim _{n \rightarrow+\infty}\left(b_{1}+b_{2}+\cdots+b_{n}\right)=\sqrt{2}-1$, find the first term and common difference of $\left\{a_{n}\right\}$.
6. Let the formula be $d$, then $d=a_{2}-a_{1}>0$, from $b_{1} b_{3}=b_{2}^{2}$, we get $a_{1}^{2}\left(a_{1}+2 d\right)^{2}=\left(a_{1}+d\right)^{4}$, simplifying to $2 a_{1}^{2}+4 a_{1} d+$ $d^{2}=0$, solving for $d=(-2 \pm \sqrt{2}) a_{1}$, hence $a_{1}<0$. If $d=(-2-\sqrt{2}) a_{1}$, then $q=\frac{a_{2}^{2}}{a_{1}^{2}}=(1+\sqrt{2})^{2}$; if $d=(-2+\sqrt{2}) a_{1}$, then $q=(\sqrt{2}-1)^{2}$, but $\lim _{n \rightarrow+\infty}\left(b_{1}+b_{2}+\cdots+b_{n}\right)=\sqrt{2}+1=\frac{b_{1}}{1-q}$ exists, so $|q|<1$. Therefore, $q=(\sqrt{2}-1)^{2}$, thus $\frac{a_{1}^{2}}{1-(\sqrt{2}-1)^{2}}=\sqrt{2}+1$, leading to $a_{1}^{2}=(2 \sqrt{2}-2)(\sqrt{2}+1)=2$, so $a_{1}=-\sqrt{2}, d=(-2+\sqrt{2}) a_{1}=$ $2 \sqrt{2}-2$