In an arithmetic sequence with a common difference of $d$, $a_{1}=1$ and $a_{n}=81$. In a geometric sequence with a common ratio of $q$, $b_{1}=1$ and $b_{n}=81$. We also know that $\frac{q}{d}=0.15$. Find all such sequences.
I. solution. Based on the formula for the $n$-th term of an arithmetic and geometric sequence:

$$
81=1+(n-1) d
$$

and

$$
81=q^{n-1}
$$

From (1),

$\mathrm{d}=\frac{80}{n-1}$, and from $\frac{q}{d}=0.15$ we get

$$
q=0.15 d=0.15 \cdot \frac{80}{n-1}=\frac{12}{n-1}
$$

Substituting this into (2) we get $81=\left(\frac{12}{n-1}\right)^{n-1}$.

Here, if $n-1 \geq 13$ then $\frac{12}{n-1} \leq 1$, so $\left(\frac{12}{n-1}\right)^{n-1} \leq 1$ would be true, and for $n=13$ we would have $\left(\frac{12}{n-1}\right)^{n-1}=1$. Therefore, $2 \leq n \leq 12$. Let's check these 11 cases:

$$
\begin{array}{llll}
\left(\frac{12}{1}\right)^{1}=12 ; & \left(\frac{12}{2}\right)^{2}=36 ; & \left(\frac{12}{3}\right)^{3}=64 ; & \left(\frac{12}{4}\right)^{4}=81 \\
\left(\frac{12}{5}\right)^{5} \approx 79.63 ; & \left(\frac{12}{6}\right)^{6}=64 ; & \left(\frac{12}{7}\right)^{7} \approx 43.51 ; & \left(\frac{12}{8}\right)^{8} \approx 25.6 \\
\left(\frac{12}{9}\right)^{9} \approx 13.32 ; & \left(\frac{12}{10}\right)^{10} \approx 6.19 ; & \left(\frac{12}{11}\right)^{11} \approx 2.60 .
\end{array}
$$

It can be seen that only for $n=5$ we get a solution, in which case $q=\frac{12}{4}=3$ and $d=\frac{80}{4}=20$.

II. solution. Based on the definition of an arithmetic sequence,

$$
(n-1) d=a_{n}-a_{1}=81-1=80
$$

so $d=\frac{80}{n-1}$. Since $\frac{q}{d}=0.15=\frac{3}{20}$, then $d=\frac{20}{3} q$, so $\frac{20}{3} q=\frac{80}{n-1}$. From this, $q=\frac{12}{n-1}$ follows. Since $n \geq 2$ and is an integer, $q$ is a positive rational number. Let $q=\frac{p}{s}$, where $(p, s)=1$. Then in the geometric sequence,

$$
81=q^{n-1}=\frac{p^{n-1}}{s^{n-1}}
$$

Since $(p, s)=1$, if $s>1$ this would not be an integer. Therefore, $s=1$, so $q$ is a positive integer. This implies that $n-1 \mid 12$ and is positive, so the possible values of $n-1$ are $1,2,3,4,6,12$. Then the values of $q$, and $q^{n-1}$ are respectively $12,6,4,3,2,1$ and $12,36,64,81,64,1$. It can be seen that only $n-1=4$ gives a good solution, in which case $d=\frac{80}{4}=20, a_{n}=1+4 \cdot 20=81$ is also satisfied. Therefore, the only solution is: $n=5, q=3, d=20$.