Mathematics Advanced • Year 12 • Module 7 • Lesson 7
Introducing Annuities, Future Value
Build procedural fluency in calculating the future value of an ordinary annuity from regular contributions.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete the future value of an ordinary annuity formula:
FV = a × ____________________________
Q1.2 Define each symbol:
a = ________________________________
r = ________________________________
n = ________________________________
Q1.3 An ordinary annuity has payments at the ____________ of each period. Total contributions over n periods = ____________.
2. Worked example, $300/month for 5 years at 4.8% p.a.
Follow each line. Reasons appear in italics on the right.
Problem. Deposit $300 at the end of every month at 4.8% p.a. compounded monthly. Find FV after 5 years.
Step 1, Convert the rate and time to match the contribution frequency.
r = 0.048 / 12 = 0.004 (monthly). n = 5 × 12 = 60 (months).
Reason: r and n must be in the same unit as the contribution.
Step 2, Substitute into the FV formula.
FV = 300 × [(1.004)⁶⁰ − 1] / 0.004
Reason: this is the closed-form sum of the GP of contributions.
Step 3, Evaluate.
(1.004)⁶⁰ = 1.27049. [(1.27049 − 1)/0.004] = 67.623.
FV = 300 × 67.623 = $20,286.90.
Step 4, Decompose into contributions + interest.
Total contributions = 300 × 60 = $18,000.
Interest earned = 20,286.90 − 18,000 = $2,286.90.
Conclusion. FV ≈ $20,286.90; about $2,287 of that is interest.
3. Faded example, fill in the missing steps
Find the future value of $200 deposited at the end of each quarter for 8 years at 5.2% p.a. compounded quarterly. 4 marks
Step 1, Match rate and time to quarters.
r = 0.052 / ____ = ____________ ; n = 8 × ____ = ____________
Step 2, Substitute into the formula.
FV = 200 × [(1 + ____)^____ − 1] / ____
Step 3, Evaluate the factor.
(1.013)^____ = ____________. Factor = (____________ − 1) / 0.013 = ____________
Step 4, Multiply by the payment.
FV = 200 × ____________ = $____________
Conclusion. The future value is $____________.
4. Graduated practice
Show every line of substitution. Round money to the nearest cent.
Foundation, match the rate to the frequency (4 questions)
| Q | Setup | r (per period), n |
|---|---|---|
| 4.1 1 | 5% p.a. compounded monthly, 4 years | r = ______ ; n = ______ |
| 4.2 1 | 6% p.a. compounded quarterly, 7 years | r = ______ ; n = ______ |
| 4.3 1 | 4.4% p.a. compounded fortnightly, 10 years | r = ______ ; n = ______ |
| 4.4 1 | 3.6% p.a. compounded half-yearly, 12 years | r = ______ ; n = ______ |
Standard, direct FV calculations (6 questions)
Show the formula line, the (1+r)^n value, and the FV to the nearest cent.
4.5 $400 at the end of each year for 12 years at 5% p.a. 2 marks
4.6 $500 at the end of each month for 1 year at 6% p.a. compounded monthly. 2 marks
4.7 $250 at the end of each fortnight for 5 years at 5.2% p.a. compounded fortnightly. 2 marks
4.8 $1,000 at the end of each year for 20 years at 6% p.a. State both the total contributions and the FV. 2 marks
4.9 $200 at the end of each month for 8 years at 4.5% p.a. compounded monthly. 2 marks
4.10 $450 at the end of each fortnight for 25 years at 6.4% p.a. compounded fortnightly (the worked super example). 2 marks
Extension, transpose and reason (2 questions)
4.11 Liam wants FV = $50,000 after 20 years at 5% p.a. compounded annually. Find the required annual contribution a to the nearest dollar. 3 marks
4.12 Show algebraically that doubling the contribution a (with r and n unchanged) exactly doubles FV, while doubling n more than doubles FV. 3 marks
5. Self-check the easy 3
Tick once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1, FV formula
FV = a × [(1 + r)ⁿ − 1] / r.
Q1.2, Symbol definitions
a = regular (end-of-period) payment; r = interest rate per period; n = number of periods.
Q1.3, End-of-period and totals
Payments at the end of each period. Total contributions = n × a (or na).
Q3, Faded example: $200/quarter, 8 yrs, 5.2% p.a.
r = 0.052 / 4 = 0.013; n = 8 × 4 = 32. FV = 200 × [(1.013)³² − 1] / 0.013. (1.013)³² = 1.51360. Factor = (1.51360 − 1)/0.013 = 39.508. FV = 200 × 39.508 = $7,901.60.
Q4.1–4.4, Rate / period matches
4.1: r = 0.05/12 ≈ 0.004167; n = 48.
4.2: r = 0.06/4 = 0.015; n = 28.
4.3: r = 0.044/26 ≈ 0.001692; n = 260.
4.4: r = 0.036/2 = 0.018; n = 24.
Q4.5
FV = 400 × [(1.05)¹² − 1]/0.05 = 400 × (1.79586 − 1)/0.05 = 400 × 15.9171 = $6,366.85.
Q4.6
r = 0.005, n = 12. FV = 500 × [(1.005)¹² − 1]/0.005 = 500 × 12.336 = $6,168.03.
Q4.7
r = 0.052/26 = 0.002, n = 130. FV = 250 × [(1.002)¹³⁰ − 1]/0.002 = 250 × 147.890 = $36,972.50 (small rounding may vary by a few dollars).
Q4.8
FV = 1,000 × [(1.06)²⁰ − 1]/0.06 = 1,000 × 36.7856 = $36,785.59. Total contributions = 20 × 1,000 = $20,000; interest = $16,785.59.
Q4.9
r = 0.00375, n = 96. FV = 200 × [(1.00375)⁹⁶ − 1]/0.00375 = 200 × 114.984 ≈ $22,996.74.
Q4.10
r = 0.064/26 = 0.002462, n = 650. FV = 450 × [(1.002462)⁶⁵⁰ − 1]/0.002462 = 450 × 1,595.0 ≈ $717,750. (Contributions = $292,500; interest ≈ $425,250.)
Q4.11, Required annual contribution
50,000 = a × [(1.05)²⁰ − 1]/0.05 = a × 33.0660. So a = 50,000 / 33.0660 ≈ $1,512 per year (to the nearest dollar).
Q4.12, Effect of doubling a vs n
Double a. FV′ = 2a × [(1+r)ⁿ − 1]/r = 2 × FV. The contribution factor a is linear, so doubling a exactly doubles FV.
Double n. New FV″ = a × [(1+r)²ⁿ − 1]/r. Since (1+r)²ⁿ = ((1+r)ⁿ)², the numerator grows quadratically (in the (1+r)ⁿ "growth factor"), not linearly, so FV″ > 2 × FV whenever r > 0. Concretely, doubling n while halving a does not preserve FV, the extra time outweighs the lost payment size.