一个等差数列的首项为 $a$，公差为 $d$，其中 $a$ 和 $d$ 均非零。该等差数列的第一、第三和第七项是一个公比为 $r$ 的等比数列中三个连续的正数项。 证明 $r=2$。 等比数列的首项是等差数列首项的十分之一。求最小的 $n$ 值，使得该等比数列前 $n$ 项的和超过该等差数列前 $2n$ 项的和。 一个球每次垂直落到水平地面上，都会反弹到它下落高度的五分之三。它最初是从距离地面 $h$ m 的高处落下的。 用 $h$ 表示球在第三次撞击地面之前所经过的路程。 证明球运动的总路程不能超过 $4h$ m。

1. Express the 1st, 3rd, and 7th terms of the arithmetic progression and use the common ratio property to form an equation.

The 1st, 3rd, and 7th terms of the arithmetic progression are $a$, $a+2d$, and $a+6d$ respectively.

Since these are consecutive terms of a geometric progression, the ratio between consecutive terms is the common ratio $r$:
$r = \frac{a+2d}{a} = \frac{a+6d}{a+2d}$

2. Solve the equation to find the relationship between the first term $a$ and common difference $d$.

Equating the ratios:
$(a+2d)^2 = a(a+6d)$
$a^2 + 4ad + 4d^2 = a^2 + 6ad$
$4d^2 - 2ad = 0$
$2d(2d - a) = 0$

It is given that $d \neq 0$, therefore we must have:
$2d - a = 0 \implies a = 2d$

3. Substitute $a = 2d$ back into the expression for $r$ to evaluate it.

Substitute $a = 2d$ into the expression for $r$:
$r = \frac{a+2d}{a}$
$r = \frac{2d+2d}{2d}$
$r = \frac{4d}{2d} = 2 \text{ (shown)}$

1. Use the common ratio $r=2$ from part (i) to find the common difference $d$ in terms of $a$.

Since the 1st and 3rd terms of the AP ($a$ and $a+2d$) are consecutive terms of a GP with $r=2$:
$\frac{a + 2d}{a} = 2$
$a + 2d = 2a \implies d = \frac{1}{2}a$

2. Express the sum of the first $n$ terms of the GP and the sum of the first $2n$ terms of the AP in terms of $a$ and $n$.

Let the first term of the GP be $g_1$. We are given $g_1 = 0.1a$.
Sum of the first $n$ terms of the GP:
$S_n = \frac{g_1(r^n - 1)}{r - 1} = \frac{0.1a(2^n - 1)}{2 - 1} = 0.1a(2^n - 1)$

Sum of the first $2n$ terms of the AP:
$S'_{2n} = \frac{2n}{2} [ 2a + (2n - 1)d ]$
$S'_{2n} = n \left[ 2a + (2n - 1)\left(\frac{a}{2}\right) \right]$
$S'_{2n} = an \left( 2 + n - \frac{1}{2} \right) = an(n + 1.5) = 0.5an(2n + 3)$

3. Set up and simplify the inequality for the sum of the GP exceeding the sum of the AP.

$S_n > S'_{2n}$
$0.1a(2^n - 1) > 0.5an(2n + 3)$
Since the given sequence terms are positive, the AP first term $a$ must be positive ($a > 0$). Dividing both sides by $0.1a$ preserves the inequality sign:
$2^n - 1 > 5n(2n + 3)$
$2^n > 10n^2 + 15n + 1$

4. Solve the inequality to find the smallest integer $n$.

Testing positive integer values for $n$:
When $n = 9$:
$2^9 = 512$ and $10(9)^2 + 15(9) + 1 = 946$
(Since $512 < 946$, the inequality is false)

When $n = 10$:
$2^{10} = 1024$ and $10(10)^2 + 15(10) + 1 = 1151$
(Since $1024 < 1151$, the inequality is false)

When $n = 11$:
$2^{11} = 2048$ and $10(11)^2 + 15(11) + 1 = 1376$
(Since $2048 > 1376$, the inequality is true)

The smallest value of $n$ is $11$.

1. Determine the distances travelled for the initial drop and the first two rebounds.

Initial drop distance: $h$

First rebound and subsequent fall (before 2nd strike): $2 \times \left(\frac{3}{5}\right)h = \frac{6}{5}h$

Second rebound and subsequent fall (before 3rd strike): $2 \times \left(\frac{3}{5}\right)^2h = \frac{18}{25}h$

2. Sum the distances to find the total distance just before the third strike.

Total distance $= h + \frac{6}{5}h + \frac{18}{25}h$

$= \left(\frac{25}{25} + \frac{30}{25} + \frac{18}{25}\right)h$

$= \frac{73}{25}h$

1. Set up an expression for the total distance travelled.

The total distance includes the initial drop $h$, plus both the upward and downward distances for all subsequent bounces. The rebound heights form a geometric progression:
$\text{Total distance} = h + 2 \left( \frac{3}{5}h + \left(\frac{3}{5}\right)^2 h + \left(\frac{3}{5}\right)^3 h + \dots \right)$

2. Calculate the sum to infinity of the rebound heights.

The series of rebound heights is an infinite geometric progression with first term $a = \frac{3}{5}h$ and common ratio $r = \frac{3}{5}$.
Using the formula for the sum to infinity $S_\infty = \frac{a}{1 - r}$:
$S_\infty = \frac{\frac{3}{5}h}{1 - \frac{3}{5}} = \frac{\frac{3}{5}h}{\frac{2}{5}} = \frac{3}{2}h$

3. Find the theoretical maximum total distance and conclude.

Substitute $S_\infty$ back into the total distance expression:
$\text{Total distance} = h + 2 \left( \frac{3}{2}h \right) = h + 3h = 4h$
Since $4h$ is the limit of the sum to infinity, the distance travelled for any finite number of bounces will always be strictly less than $4h$. Therefore, the total distance cannot exceed $4h$ m.