IB Mathematics AI SL Amortization and annuities using technology Study Notes - New Syllabus

Solution:

 The $\$2000$ deposited each year earns interest for a different number of years:

• The first $2000 is invested for 7 years: \(2000(1.05)^7\)
• The second $2000 is invested for 6 years: \(2000(1.05)^6\)
• …
• The last $2000 is invested for 0 years: \(2000\)

So the total future value is:

$FV = 2000(1.05)7 + 2000(1.05)6 + 2000(1.05)5 + … + 2000$

This is a geometric series where:

First term \(u = 2000\),

Common ratio \(r = 1.05\),

Number of terms \(n = 8\).

$S_n = u \cdot \frac{r^n – 1}{r – 1} = 2000 \cdot \frac{(1.05)^8 – 1}{0.05} \approx 2000 \cdot 9.549 = 19098.59 $

Final value of the annuity $≈ \$19098.59$

Solution:

\(P = \frac{(0.06/12) \times 10,000}{1 – (1 + 0.06/12)^{-60}} \approx \$193.33\)
Total Paid: \(60 \times 193.33 = \$11,599.80\)
Total Interest: \(11,599.80 – 10,000 = \$1,599.80\)