一家小型贸易公司在 2000 年获得了 $250000 的利润。该公司考虑了两种不同的计划（计划 A 和计划 B）来增加其利润。 在计划 A 下，年利润每年将比前一年增加 5%。对于计划 A，求：

1. Identify the type of sequence, its first term, and its common ratio.

The annual profit increases by $5\%$ each year, which forms a geometric progression.
First term (year 2000), $a = 250000$
Common ratio, $r = 1 + \frac{5}{100} = 1.05$

2. Determine the term number corresponding to the year 2008.

Year 2000 is the 1st term ($n=1$).
Therefore, year 2008 is the 9th term ($n=9$).

3. Calculate the 9th term using the formula for the $n$th term of a geometric progression.

$u_n = a r^{n-1}$
$u_9 = 250000 \times (1.05)^{9-1}$
$u_9 = 250000 \times 1.05^8$
$u_9 \approx 369363.86$
$u_9 \approx 369000 \text{ (to 3 s.f.)}$

1. Identify the parameters for the sum of the geometric progression.

The annual profit follows a geometric progression.
For the 10 years 2000 to 2009 inclusive, the number of terms is $n = 10$.
First term, $a = 250000$
Common ratio, $r = 100\% + 5\% = 1.05$

2. Apply the geometric series sum formula to calculate the total profit.

$S_{10} = \frac{a(r^n - 1)}{r - 1}$
$S_{10} = \frac{250000(1.05^{10} - 1)}{1.05 - 1}$
$S_{10} = \frac{250000(1.628895 - 1)}{0.05}$
$S_{10} \approx 3144473.13$
Total profit $\approx \char36{}3 144 473$ (to nearest dollar)

1. Calculate the total profit over the 10 years for Plan A (from part ii).

$S_{10, A} = \frac{250000(1.05^{10} - 1)}{1.05 - 1}$\n$S_{10, A} \approx 3144473.134$

2. Formulate the total profit over the 10 years for Plan B as the sum of an arithmetic series.

$S_{10, B} = \frac{10}{2} [2(250000) + (10 - 1)D]$\n$S_{10, B} = 5(500000 + 9D)$\n$S_{10, B} = 2500000 + 45D$

3. Equate the total profits from both plans and solve for $D$.

$2500000 + 45D = 3144473.134$\n$45D = 644473.134$\n$D \approx 14321.63$