一套特别的俄罗斯套娃由七个娃娃组成。这七个娃娃的高度呈等差数列，公差为 $d$。如果最高的娃娃的高度是矮的最矮的娃娃高度的四倍，且这套娃娃中所有娃娃的高度之和为 70 cm，求 $d$。 鲍勃从平台上走下，垂直下落，并在蹦床上反弹。任何特定一次反弹与他下一次反弹之间的时间间隔是该次反弹与前一次反弹之间时间间隔的 90%。他的第一次和第二次反弹之间的时间间隔是 4 秒。 已知第 $k$ 次反弹和第 $(k + 1)$ 次反弹之间的时间间隔是第一个小于 0.4 秒的此类间隔，求 $k$。此外，求从第一次反弹到第 $k$ 次反弹的总时间，结果精确到最接近的秒。

1. Formulate an equation relating the heights of the shortest and tallest dolls.

Let $a$ be the height of the shortest doll. Since there are 7 dolls, the height of the tallest doll is the 7th term, $u_7 = a + 6d$.
Given that the tallest doll is four times the height of the shortest one:
$a + 6d = 4a$
$3a = 6d \implies a = 2d$

2. Formulate an equation using the sum of the heights.

The sum of the heights of the 7 dolls is 70 cm. Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$:
$\frac{7}{2}(2a + 6d) = 70$
$7(a + 3d) = 70$
$a + 3d = 10$

3. Solve the simultaneous equations to find the common difference $d$.

Substitute $a = 2d$ into the simplified sum equation:
$2d + 3d = 10$
$5d = 10$
$d = 2$

1. Express the time interval between bounces as a geometric progression.

Let $T_n$ be the time interval between the $n$-th and $(n+1)$-th bounce.
$T_n$ forms a geometric progression with first term $a = 4$ and common ratio $r = 0.9$.
$T_n = a r^{n-1} = 4(0.9)^{n-1}$

2. Set up the inequality for the $k$-th interval.

We are given that the interval between the $k$-th and $(k+1)$-th bounce is less than $0.4\text{ s}$.
$T_k < 0.4$
$4(0.9)^{k-1} < 0.4$
$(0.9)^{k-1} < 0.1$

3. Solve for $k$ using logarithms.

Taking natural logarithms on both sides:
$(k-1) \ln 0.9 < \ln 0.1$
Since $\ln 0.9$ is negative, the inequality sign flips when dividing:
$k - 1 > \frac{\ln 0.1}{\ln 0.9}$
$k - 1 > 21.854$
$k > 22.854$
Since $k$ must be an integer, $k = 23$.

4. Identify the number of intervals to sum for the total time.

The total time from the 1st bounce to the $k$-th bounce ($k=23$) is the sum of the first $k-1$ intervals, from $T_1$ up to $T_{22}$.
$\text{Total time} = S_{22} = \frac{a(1-r^{22})}{1-r}$

5. Calculate the total time and round to the nearest second.

$S_{22} = \frac{4(1-0.9^{22})}{1-0.9}$
$S_{22} = \frac{4(1-0.098477)}{0.1}$
$S_{22} \approx 36.061 \text{ s}$
To the nearest second, the total time is $36\text{ s}$.