The sum of the first $n$ terms of a series is given by $S_n = 3n(n + 2)$. Show that this series follows an arithmetic progression. The second, seventh and $m$th term of the series in part (a) are the first three consecutive terms of a geometric series, such that its $n$th term is given by $v_n$. Find the value of $m$ and explain whether the sum to infinity of the geometric series exists. The $n$th term of another geometric series is given by $w_n = \text{e}^{5 + nx(x + 1)}$, where $x$ is a constant. Find the range of values of $x$ such that this geometric series converges. Using $x = -0.5$ in part (c) and given that the sum of the first $n$ terms of the geometric series in part (b) first exceeds the $n$th term of the geometric series in part (c), i.e., $\left(\sum_{r=1}^{n} v_r\right) > w_n$, determine the least value of $n$.

1. State the relationship between the $n$th term and the sum of the first $n$ terms.

$u_n = S_n - S_{n-1}$

2. Substitute the given expression for $S_n$ to find $u_n$ in terms of $n$.

$u_n = 3n(n + 2) - 3(n - 1)(n - 1 + 2)$
$u_n = 3n^2 + 6n - 3(n - 1)(n + 1)$
$u_n = 3n^2 + 6n - 3(n^2 - 1)$
$u_n = 3n^2 + 6n - 3n^2 + 3$
$u_n = 6n + 3$

3. Calculate the difference between consecutive terms to show it is a constant.

$u_n - u_{n-1} = (6n + 3) - [6(n - 1) + 3]$
$u_n - u_{n-1} = 6n + 3 - (6n - 6 + 3)$
$u_n - u_{n-1} = 6n + 3 - (6n - 3)$
$u_n - u_{n-1} = 6$

Since the difference between consecutive terms is a constant ($6$) that does not depend on $n$, the series follows an arithmetic progression.

1. Find the general term of the arithmetic series and compute the 2nd, 7th, and $m$th terms.

Using the relationship $u_n = S_n - S_{n-1}$ (typically found in part (a)):
$u_n = 3n(n+2) - 3(n-1)(n+1)$
$u_n = 3n^2 + 6n - 3(n^2 - 1) = 6n + 3$

Hence, the required terms are:
$u_2 = 6(2) + 3 = 15$
$u_7 = 6(7) + 3 = 45$
$u_m = 6m + 3$

2. Find the common ratio of the geometric series and solve for $m$.

The 2nd, 7th, and $m$th terms of the AP are consecutive terms of a geometric series.
The common ratio $r$ is:
$r = \frac{u_7}{u_2} = \frac{45}{15} = 3$

Since $u_m$ is the next consecutive term in the geometric series:
$u_m = u_7 \times r$
$6m + 3 = 45 \times 3$
$6m + 3 = 135$
$6m = 132 \implies m = 22$

3. Explain whether the sum to infinity exists.

For a geometric series to have a sum to infinity, it must converge, meaning its common ratio $r$ must satisfy $|r| < 1$.
Since $r = 3 \ge 1$, the geometric series diverges and its sum to infinity does not exist.

1. Determine the common ratio of the geometric series.

The common ratio $r$ is given by the ratio of consecutive terms:
$r = \frac{w_{n+1}}{w_n} = \frac{\text{e}^{5 + (n+1)x(x + 1)}}{\text{e}^{5 + nx(x + 1)}}$
$r = \text{e}^{5 + nx(x + 1) + x(x + 1) - (5 + nx(x + 1))}$
$r = \text{e}^{x(x + 1)}$

2. State the condition for convergence of an infinite geometric series.

For a geometric series to converge, its common ratio $r$ must satisfy $|r| < 1$, which means $-1 < r < 1$.
$-1 < \text{e}^{x(x + 1)} < 1$

3. Solve the inequality to find the range of values for $x$.

Since $\text{e}^{x(x + 1)} > 0$ for all real $x$, the condition $-1 < \text{e}^{x(x + 1)}$ is naturally satisfied. We only need to solve:
$\text{e}^{x(x + 1)} < 1$
$\text{e}^{x(x + 1)} < \text{e}^0$
Equating the exponents:
$x(x + 1) < 0$
The critical values of the quadratic expression are $x = 0$ and $x = -1$. Because the coefficient of $x^2$ is positive (a "U-shape" parabola), the expression is less than zero between its roots:
$-1 < x < 0$

1. Set up the inequality using expressions for the sum of $v_n$ and the $n$th term $w_n$.

From part (b), the geometric series $v_n$ has first term $a = 15$ and common ratio $r = 3$.
The sum of the first $n$ terms is:
$\sum_{r=1}^{n} v_r = \frac{15(3^n - 1)}{3 - 1} = 7.5(3^n - 1)$
Substitute $x = -0.5$ into $w_n$:
$w_n = \text{e}^{5 + n(-0.5)(-0.5 + 1)} = \text{e}^{5 - 0.25n}$
The inequality to solve is:
$7.5(3^n - 1) > \text{e}^{5 - 0.25n}$

2. Evaluate the inequality for integer values of $n$ to find the minimum $n$.

Testing positive integer values of $n$ (using a graphing calculator table or manual evaluation):
When $n = 1$:
LHS $= 7.5(3^1 - 1) = 15$
RHS $= \text{e}^{4.75} \approx 115.58$

When $n = 2$:
LHS $= 7.5(3^2 - 1) = 60$
RHS $= \text{e}^{4.5} \approx 90.02$

When $n = 3$:
LHS $= 7.5(3^3 - 1) = 195$
RHS $= \text{e}^{4.25} \approx 70.11$

Since $195 > 70.11$, the inequality is first satisfied when $n = 3$.