An arithmetic progression has first term $a$ and common difference $d$, where $a$ and $d$ are non-zero. The first, third and seventh terms of the arithmetic progression are three consecutive positive terms of a geometric progression with common ratio $r$. Show that $r=2$. The first term of the geometric progression is one-tenth that of the first term of the arithmetic progression. Find the smallest value of $n$ such that the sum of the first $n$ terms of the geometric progression exceeds the sum of the first $2n$ terms of the arithmetic progression. Each time that a ball falls vertically on to a horizontal floor, it rebounds to three-fifth of the height from which it fell. It is initially dropped from a point $h$ m above the floor. Find the distance travelled by the ball just before it strikes the floor for the third time in terms of $h$. Show that the total distance travelled by the ball cannot exceed $4h$ 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.