A financial institution, Future Investments Inc., has introduced a new investment scheme. The scheme pays a compound interest of 5% per annum at the end of the year, based on the amount in the account at the beginning of each year. John and Sarah are both interested in this scheme.

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 $26$.