Source: ACE Exam Paper 1 ADV
Tags: arithmetic sequence, geometric series, common difference, common ratio, annuity, present value, future value, compound interest, recurrence relation, sum of series
This section covers arithmetic and geometric sequences (finding terms, summing), financial applications (annuities, compound interest, loan recurrence relations), and reading annuity tables. Most questions follow directly from the standard formulas, so knowing which formula to reach for is the key skill.
Arithmetic sequence
A sequence where the difference between consecutive terms is constant. General term: Tₙ = a + (n - 1)d, where a is the first term and d is the common difference.
Common difference (d)
The constant difference between consecutive terms in an arithmetic sequence. Found by subtracting any term from the next: d = T₂ - T₁.
Geometric sequence
A sequence where the ratio between consecutive terms is constant. General term: Tₙ = arⁿ⁻¹, where a is the first term and r is the common ratio.
Common ratio (r)
The constant ratio between consecutive terms: r = T₂/T₁.
Annuity
A series of equal payments made at regular intervals. The present value is what the annuity is worth now; the future value is what it will be worth at the end.
Present value of an annuity
The lump sum today that is equivalent to the stream of future payments, discounted at a given interest rate.
Future value of an annuity
The total accumulated value of all payments plus compounded interest at the end of the investment period.
Recurrence relation
A formula that defines each term in a sequence using the previous term. For loans: Vₙ₊₁ = Vₙ × (1 + r) - M, where r is the periodic interest rate and M is the periodic payment.
Compound interest
Interest calculated on the principal plus previously accumulated interest. A = P(1 + r)ⁿ.
First term a = 4, common difference d = 5.
(a) nth term rule:
Tₙ = 4 + (n - 1)(5) = 5n - 1
(b) 25th term:
T₂₅ = 5(25) - 1 = 124
(c) Sum of first 100 terms:
S₁₀₀ = (100/2)[2(4) + (99)(5)] = 50[8 + 495] = 50 × 503 = 25 150
The 3rd term is 1.25 and the 7th term is 20.
T₃ = ar² = 1.25
T₇ = ar⁶ = 20
Dividing: ar⁶ / ar² = r⁴ = 20/1.25 = 16, so r = 2.
From ar² = 1.25: a(4) = 1.25, so a = 0.3125 (or 5/16).
Florence invests $1000 at 5% compounded annually.
(a) Amount after 100 years (single deposit):
A = 1000(1.05)¹⁰⁰
Using logarithms or a calculator: (1.05)¹⁰⁰ ≈ 131.501. So A ≈ $131 501.
(b) With an additional $1000 at the start of each subsequent year (annuity-due):
The first $1000 grows for 100 years, the second for 99 years, and so on down to the last $1000 growing for 1 year.
Total = 1000(1.05)¹⁰⁰ + 1000(1.05)⁹⁹ + ... + 1000(1.05)¹
This is a geometric series with first term 1000(1.05), common ratio 1.05, and 100 terms.
S = 1000(1.05) × [(1.05)¹⁰⁰ - 1] / (1.05 - 1) = 1050 × [131.501 - 1] / 0.05 = 1050 × 130.501 / 0.05 = 1050 × 2610.025 ≈ $2 740 526
$12,000 contributed per year for 6 years at 3% per annum.
From the table, the present value factor for 6 years at 3% is 5.4172.
Present value = 12,000 × 5.4172 = $65 006.40
Answer: (C).
(a) $32,000/year at 8% for 4 years:
From the table, the future value factor for 4 years at 8% is 4.51.
Future value = 32,000 × 4.51 = $144 320
(b) $6,300 every 6 months at 8% p.a. compounded biannually for 2 years:
The periodic rate is 8%/2 = 4% per half-year. There are 4 half-year periods.
From the table, the future value factor for 4 periods at 4% is 4.25.
Future value = 6,300 × 4.25 = $26 775
Loan of $58,000 at 6% p.a. compounded monthly, with $810 monthly payments.
Monthly interest rate: 6%/12 = 0.5% = 0.005.
The recurrence relation is:
Vₙ₊₁ = Vₙ × 1.005 - 810, where V₀ = 58 000
A = 10e⁻ᵏᵗ, where A is the undissolved amount in kg after t hours.
(a) Finding k when A = 3.6 at t = 5:
3.6 = 10e⁻⁵ᵏ e⁻⁵ᵏ = 0.36 -5k = ln(0.36) k = -ln(0.36)/5 = 1.0217.../5 ≈ 0.204 (to 3 d.p.)
(b) Time for 1 kg to remain:
1 = 10e⁻⁰·²⁰⁴ᵗ e⁻⁰·²⁰⁴ᵗ = 0.1 -0.204t = ln(0.1) t = -ln(0.1)/0.204 = 2.3026/0.204 ≈ 11.3 hours (to 1 d.p.)
Arithmetic: Tₙ = a + (n - 1)d
Arithmetic sum: Sₙ = (n/2)[2a + (n - 1)d] = (n/2)(a + l), where l is the last term
Geometric: Tₙ = arⁿ⁻¹
Geometric sum: Sₙ = a(rⁿ - 1)/(r - 1) for r ≠ 1
Compound interest: A = P(1 + r)ⁿ
Loan recurrence: Vₙ₊₁ = Vₙ(1 + r) - M
Exponential decay: A = A₀e⁻ᵏᵗ
⚠️ Arithmetic vs geometric: check whether the sequence has a constant difference (arithmetic) or a constant ratio (geometric) before choosing a formula.
⚠️ For annuity table questions, make sure you read the correct row (number of periods) and column (interest rate). A misread cell gives a plausible but wrong answer.
⚠️ Monthly compounding means the annual rate must be divided by 12, and the number of periods is years × 12.
⚠️ "Compounded biannually" means twice per year (every 6 months), not every 2 years. Divide the annual rate by 2 and multiply the years by 2 for the number of periods.
⚠️ In geometric series problems where you are given two terms, divide one by the other to eliminate a. This gives you rⁿ directly.
⚠️ For exponential decay, remember that k is positive and the exponent is -kt. If you get a negative k, check your signs.
Q: For the arithmetic sequence 4, 9, 14, 19, ..., find the 25th term.
A: Tₙ = 5n - 1. T₂₅ = 124.
Q: Find the sum of the first 100 terms of the sequence 4, 9, 14, 19, ...
A: S₁₀₀ = 50[8 + 495] = 25 150.
Q: The 3rd and 7th terms of a geometric sequence are 1.25 and 20. Find the first term.
A: r⁴ = 16, so r = 2. Then a = 1.25/4 = 0.3125.
Q: Write a recurrence relation for a $58,000 loan at 6% p.a. compounded monthly with $810 monthly payments.
A: Vₙ₊₁ = 1.005Vₙ - 810, V₀ = 58 000.
Q: Present value of $12,000/year for 6 years at 3% (PV factor = 5.4172)?
A: 12,000 × 5.4172 = $65,006.40.
Q: A = 10e⁻ᵏᵗ, and A = 3.6 when t = 5. Find k to 3 d.p.
A: k = -ln(0.36)/5 ≈ 0.204.
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