一家名为“未来投资公司”（Future Investments Inc.）的金融机构推出了一项新的投资计划。该计划在每年年底支付 5% 的年复利，利息基于每年年初账户中的金额计算。 约翰（John）和莎拉（Sarah）都对这个计划很感兴趣。

1. Multiply the initial investment by the interest multiplier to find the amount at the end of the first year.

\text{Amount} = x \times (1 + 0.05) = 1.05x

1. Establish the pattern of the total amount at the end of each year to form a series.

At the end of year 1: $1.05x$
At the end of year 2: $(1.05x + x)(1.05) = 1.05^2x + 1.05x$
At the end of year $3$: $(1.05^2x + 1.05x + x)(1.05) = 1.05^3x + 1.05^2x + 1.05x$
At the end of year $n$, the total amount is:
$1.05x + 1.05^2x + 1.05^3x + \dots + 1.05^nx$

2. Identify the parameters of the geometric progression.

This expression is the sum of a geometric progression with:
First term, $a = 1.05x$
Common ratio, $r = 1.05$
Number of terms $= n$

3. Apply the sum of geometric series formula and simplify to obtain the final expression.

$S_n = \frac{a(r^n - 1)}{r - 1}$
$S_n = \frac{1.05x(1.05^n - 1)}{1.05 - 1}$
$S_n = \frac{1.05x(1.05^n - 1)}{0.05}$
$S_n = 21x(1.05^n - 1) \text{ (shown)}$

1. Set up the inequality for the amount at the end of $n$ years using the formula from (a)(ii).

$21(10\,000)(1.05^n - 1) > 500\,000$
$210\,000(1.05^n - 1) > 500\,000$
$1.05^n - 1 > \frac{50}{21}$
$1.05^n > \frac{71}{21}$

2. Solve the exponential inequality using logarithms.

$n \ln(1.05) > \ln\left(\frac{71}{21}\right)$
$n > \frac{\ln(71/21)}{\ln(1.05)}$
$n > 24.967$

3. Determine the amount in the account around the 25th year to check exactly when it exceeds $500,000.

Amount at the end of the 24th year:
$S_{24} = 210\,000(1.05^{24} - 1) \approx \$467\,270.99$

Amount at the start of the 25th year (after John's $\char36{}10\,000$ deposit):
$\$467\,270.99 + \$10\,000 = \$477\,270.99$
Since $\char36{}477\,270.99 < \char36{}500\,000$, the total has not yet exceeded $\char36{}500\,000$.

Amount at the end of the 25th year:
$S_{25} = 210\,000(1.05^{25} - 1) \approx \$501\,134.54$

4. Conclude the final answer based on the context.

Since $\char36{}501\,134.54 > \char36{}500\,000$, the total first exceeds $\char36{}500\,000$ at the end of the 25th year.

1. Identify the amount invested at the start of the $r$-th year.

The amounts Sarah invests at the start of each year form an arithmetic progression with first term $6000$ and common difference $400$.
Amount invested at the start of the $r$-th year = $6000 + 400(r-1)$.

2. Determine the time period and future value of the $r$-th investment.

The investment made at the start of the $r$-th year remains in the account until the end of the $n$-th year, accumulating interest for $n - r + 1 = n - (r - 1)$ full years.
The accumulated value of this single investment at the end of $n$ years is:
$[6000 + 400(r-1)](1.05)^{n-(r-1)}$

3. Construct the total sum and compare it to find $a$ and $b$.

The total amount in the account is the sum of the accumulated values of all investments from the 1st year to the $n$-th year:
$\text{Total amount} = \sum_{r=1}^n [6000 + 400(r-1)](1.05)^{n-(r-1)}$
Comparing this with the given expression $\sum_{r=1}^n [a + b(r-1)](1.05)^{n-(r-1)}$, we obtain:
$a = 6000$
$b = 400$

1. Set up the inequality for the amount at the end of $n$ years.

Using the expression from (b)(i) with $a=6000$ and $b=400$, we require the amount at the end of $n$ complete years to exceed $500 000:
$\sum_{r=1}^n [6000 + 400(r-1)](1.05)^{n-r+1} > 500 000$

2. Use a graphing calculator to evaluate the sum for integer values of $n$ and determine the answer.

Using a graphing calculator to tabulate the values of the sum for integer values of $n$:
When $n = 25$, Amount $\approx 491\,588.36$
When $n = 26$, Amount $\approx 532\,967.77$

Since the end-of-year amount first exceeds $500 000 when $n = 26$, the number of complete years is $26$.