A finite sequence $\{a_n\}$ has 50 terms and is such that $a_{n+1} = a_n + 0.15$ for $n = 1,2,3,\ldots,49$. Another infinite sequence $\{b_m\}$ is such that $b_1 = a_{50}$ and $\dfrac{b_m}{b_{m-1}} = 0.98$ for $m \ge 2$.

1. Express the 50th term in terms of the first term.

The given recurrence relation $a_{n+1} - a_n = 0.15$ indicates that the sequence is an arithmetic progression with common difference $d = 0.15$.
Using the formula for the $n$-th term $a_n = a_1 + (n-1)d$,
$a_{50} = a_1 + (50 - 1)(0.15)$
$a_{50} = a_1 + 49(0.15) = a_1 + 7.35$

2. Substitute into the given equation and solve for $a_1$.

Given that $a_{50} = 99 a_1$, we substitute the expression for $a_{50}$:
$a_1 + 7.35 = 99 a_1$
$98 a_1 = 7.35$
$a_1 = \frac{7.35}{98}$
$a_1 = 0.075 \text{ (shown)}$

1. Set up the sum of the first 50 terms using the arithmetic progression formula.

$\sum_{n=1}^{50} a_n = \frac{n}{2}(a_1 + a_{n})$
$S_{50} = \frac{50}{2}(a_1 + a_{50})$

2. Substitute the relationship $a_{50} = 99a_1$ from part (i) to efficiently evaluate the sum without a calculator.

$S_{50} = 25(a_1 + 99a_1)$
$S_{50} = 25(100a_1)$
Since $a_1 = 0.075$:
$S_{50} = 2500(0.075) = 187.5$

1. Calculate the value of $a_{25}$.

The sequence $\{a_n\}$ is an arithmetic progression with first term $a_1 = 0.075$ and common difference $d = 0.15$.
$a_{25} = a_1 + 24d$
$a_{25} = 0.075 + 24(0.15) = 3.675$

2. Set up the inequality $b_k < a_{25}$.

The sequence $\{b_m\}$ is a geometric progression with first term $b_1 = a_{50} = 99(0.075) = 7.425$ and common ratio $r = 0.98$.
Using the formula for the $k$th term of a GP, we have $b_k = 7.425(0.98)^{k-1}$.
$7.425(0.98)^{k-1} < 3.675$
$(0.98)^{k-1} < \frac{3.675}{7.425}$
$(0.98)^{k-1} < \frac{49}{99}$

3. Solve the inequality for $k$.

Take the natural logarithm on both sides. Remember that $\ln(0.98)$ is negative, so dividing by it reverses the inequality sign:
$(k-1)\ln(0.98) < \ln\left(\frac{49}{99}\right)$
$k-1 > \frac{\ln(49/99)}{\ln(0.98)}$
$k-1 > 34.823\dots$
$k > 35.823\dots$
Since $k$ must be an integer, the smallest value is $k = 36$.

1. Formulate the inequality comparing the sum of the first $h$ terms to 99% of the sum to infinity

Let $S_h$ be the sum of the first $h$ terms and $S_\infty$ be the sum to infinity.

$S_h > 0.99 S_\infty$

$\frac{b_1(1 - 0.98^h)}{1 - 0.98} > 0.99 \left( \frac{b_1}{1 - 0.98} \right)$

2. Simplify the inequality to isolate the exponential term

Since $b_1 > 0$ (as $b_1 = a_{50} = 7.425$) and $1-0.98 > 0$, we can divide both sides by $\frac{b_1}{1-0.98}$:

$1 - 0.98^h > 0.99$
$0.98^h < 0.01$

3. Solve for $h$ using logarithms

Taking the natural logarithm on both sides:

$\ln(0.98^h) < \ln(0.01)$
$h \ln(0.98) < \ln(0.01)$

Since $\ln(0.98)$ is a negative value, the inequality sign flips when dividing:
$h > \frac{\ln(0.01)}{\ln(0.98)}$
$h > 227.95$

Since $h$ must be an integer, the least value of $h$ is 228.

1. Find the first term and common ratio of the new series.

From part (i), $a_1 = 0.075$, so $b_1 = a_{50} = 99(0.075) = 7.425$.
The sequence of terms is $b_1, b_4, b_7, \ldots$ which corresponds to $m = 0, 1, 2, \ldots$
This is a geometric series with the first term $B_1 = b_1 = 7.425$.
Since $\frac{b_k}{b_{k-1}} = r = -0.98$, the common ratio of this new series is:
$R = \frac{b_4}{b_1} = r^3 = (-0.98)^3 = -0.941192$

2. Calculate the sum to infinity of the new series.

Since $|R| = |-0.941192| < 1$, the infinite series converges. Using the sum to infinity formula:
$\begin{aligned} \sum_{m=0}^{\infty} b_{1+3m} &= \frac{B_1}{1 - R} \\ &= \frac{7.425}{1 - (-0.941192)} \\ &= \frac{7.425}{1.941192} \\ &\approx 3.82497 \\ &\approx 3.82 \text{ (to 3 s.f.)} \end{aligned}$