A particular set of Russian dolls consists of seven dolls. The heights of the seven dolls are in arithmetic progression with common difference $d$. If the height of the tallest doll is four times the height of the shortest one, and the sum of the heights of all the dolls in the set is 70 cm, find $d$. Bob stepped off a platform, fell vertically, and bounced on a trampoline. The time interval between any particular bounce and his next bounce is 90% of the time interval between that particular bounce and the preceding bounce. The interval between his first and second bounces is 4 seconds. Given that the interval between the $k$ th bounce and the $(k + 1)$ th bounce is the first such interval less than 0.4 seconds, find $k$. Also, find the total time from the first bounce to the $k$ th bounce, giving your answer to the nearest second.

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 $36\text{ s}$.