等差数列的首项为 $a$，公差为 $d$，其中 $a$ 和 $d$ 均不为零。该等差数列的第一、第三和第七项是一个公比为 $r$ 的等比数列的三个连续正数项。 证明 $r=2$。 等比数列的首项是该等差数列首项的十分之一。求最小的 $n$ 值，使得该等比数列的前 $n$ 项和超过该等差数列的前 $2n$ 项和。 每次球垂直落到水平地面上时，它都会反弹到下落高度的五分之三。最初它从距离地面 $h$ 米高处落下。 用 $h$ 表示球在第三次撞击地面之前所经过的路程。 证明球所运动的总路程不能超过 $4h$ 米。

1. Express the given AP terms and set up an equation using the property of a geometric progression

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

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

2. Solve the equation to find the relationship between $a$ and $d$

Cross-multiplying gives:
$(a+2d)^2 = a(a+6d)$
$a^2 + 4ad + 4d^2 = a^2 + 6ad$
$4d^2 - 2ad = 0$
$2d(2d - a) = 0$

Since it is given that $d \neq 0$, we have:
$2d - a = 0 \implies a = 2d$

3. Substitute the relationship back to show $r=2$

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

1. Establish the relationship between $a$ and $d$ and state the sign of $a$.

From part (a)(i), the common ratio of the geometric progression is $r = 2$.
The first and third terms of the AP are $a$ and $a + 2d$ respectively. Since they are consecutive terms of the GP:
$\frac{a + 2d}{a} = 2 \implies a + 2d = 2a \implies d = \frac{1}{2}a$
Because these terms are given to be positive, the first term $a$ must be positive ($a > 0$).

2. Formulate expressions for the sums of the GP and the AP.

For the GP, the first term is given as $\frac{1}{10}a$ and $r = 2$. The sum of the first $n$ terms is:
$S_{\text{GP}} = \frac{\frac{a}{10}(2^n - 1)}{2 - 1} = \frac{a}{10}(2^n - 1)$
For the AP, using $d = \frac{1}{2}a$, the sum of the first $2n$ terms is:
$S_{\text{AP}} = \frac{2n}{2} \left[2a + (2n - 1)d\right] = n \left[2a + (2n - 1)\left(\frac{1}{2}a\right)\right]$
$S_{\text{AP}} = na \left(2 + n - \frac{1}{2}\right) = na(n + 1.5)$

3. Set up and simplify the inequality.

We require $S_{\text{GP}} > S_{\text{AP}}$:
$\frac{a}{10}(2^n - 1) > na(n + 1.5)$
Since $a > 0$, we can divide both sides by $a$ without changing the inequality sign:
$\frac{1}{10}(2^n - 1) > n^2 + 1.5n$
Multiply both sides by 10:
$2^n - 1 > 10n^2 + 15n$
$2^n > 10n^2 + 15n + 1$

4. Solve for the smallest integer $n$.

We test integer values for $n$ to find the smallest value that satisfies the inequality.
When $n = 10$:
LHS $= 2^{10} = 1024$
RHS $= 10(10)^2 + 15(10) + 1 = 1151$
$1024 > 1151$ is False.

When $n = 11$:
LHS $= 2^{11} = 2048$
RHS $= 10(11)^2 + 15(11) + 1 = 1210 + 165 + 1 = 1376$
$2048 > 1376$ is True.

Thus, the smallest value of $n$ is $11$.

1. Determine the distance travelled during the initial drop, first rebound, and second rebound.

Distance of initial drop $= h$
Distance of first rebound (up and down) $= 2 \left(\frac{3}{5}h\right) = \frac{6}{5}h$
Distance of second rebound (up and down) $= 2 \left(\frac{3}{5}\right)^2 h = \frac{18}{25}h$

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

Total distance $= h + \frac{6}{5}h + \frac{18}{25}h$
$= \frac{25h + 30h + 18h}{25}$
$= \frac{73}{25}h$

1. Express the total distance travelled as an infinite series.

The total distance travelled involves the initial drop and two times the height of each subsequent rebound (one upward movement and one downward movement).
$\text{Total distance} = h + 2\left(\frac{3}{5}h\right) + 2\left(\frac{3}{5}\right)^2 h + 2\left(\frac{3}{5}\right)^3 h + \dots$

2. Identify the infinite geometric progression and set up the sum to infinity.

The terms after the initial drop form an infinite geometric progression with first term $a = 2\left(\frac{3}{5}h\right) = \frac{6}{5}h$ and common ratio $r = \frac{3}{5}$.
The sum to infinity is given by $S_\infty = \frac{a}{1 - r}$.
$S_\infty = \frac{\frac{6}{5}h}{1 - \frac{3}{5}}$

3. Evaluate the sum to infinity and conclude the proof.

$S_\infty = \frac{\frac{6}{5}h}{\frac{2}{5}} = 3h$
$\text{Total maximum distance} = h + 3h = 4h$
Since the sum to infinity represents the limiting value of the total distance travelled, and the sequence of finite sums is strictly increasing, the total distance for any finite number of bounces will always be strictly less than $4h$. Therefore, the total distance cannot exceed $4h$ m.