Form 5 maths・Non-foundation topic

Arithmetic and geometric sequences

Q: How can infinitely many numbers add to a finite number?

Each piece is smaller than the last, so you get closer and closer to the total but never pass it, like walking half the remaining distance to a wall again and again.

Q: How do I know if it's AS or GS?

Subtract neighbours: same difference → AS. Divide neighbours: same ratio → GS.

Q: Why do we check n and n − 1?

To be sure that n is the first value that works, not just a value that works.

Arithmetic and geometric sequences, their general terms and sums, and the sum to infinity. The unit is NF and is one of the harder long-question topics.

What you will learn

7.1–7.2 Arithmetic sequences (AS): properties and general term. Tₙ = ½(Tₙ₋₁ + Tₙ₊₁); kTₙ + a is also an AS.
7.3–7.4 Geometric sequences (GS): properties and general term. Tₙ² = Tₙ₋₁ × Tₙ₊₁; kTₙ is also a GS.
7.5 Sums of a finite number of terms of an AS and a GS.
7.6 Sum to infinity of a GS.
7.7 Real-life problems: interest, growth, depreciation.

You need first: Number and algebra basics, Percentages, Inequalities and linear programming, Exponential and logarithmic functions

Key terms

English	中文
arithmetic sequence	等差數列
geometric sequence	等比數列
first term	首項
common difference	公差
common ratio	公比
general term, nth term	通項，第 n 項
sum of the first n terms	首 n 項之和
sum to infinity	無限項之和

Standard steps

The formula card

	AS (add d each time)	GS (multiply by r each time)
general term	Tₙ = a + (n − 1)d	Tₙ = arⁿ⁻¹
sum of n terms	Sₙ = n/2 [2a + (n − 1)d] = n/2 (a + l)	Sₙ = a(rⁿ − 1)/(r − 1), r ≠ 1
sum to infinity	—	S∞ = a/(1 − r), only if −1 < r < 1
test	T₂ − T₁ = T₃ − T₂	T₂/T₁ = T₃/T₂

Why n − 1: to get from the 1st term to the nth term you take n − 1 steps.

Find a and d (or a and r)

Turn each given fact into an equation using the formula card.
Two facts give two equations. Solve simultaneously.
- For a GS, divide one equation by the other to cancel a.
Substitute back to find the other unknown, then answer the question (Tₙ or Sₙ).

"Least n such that…"

Write the sum or term as an inequality, for example Sₙ > 1000.
AS: this is a quadratic inequality in n. Solve the quadratic, then take the next whole number.
GS: rⁿ appears. Take logs: n log r > … (if 0 < r < 1, log r is negative, so the sign flips).
Check the answer n and n − 1 in the original inequality.

Worked examples

Example 1: AS: a = 5, d = 3. Find T₂₀ and S₂₀.

T₂₀ = 5 + 19(3) = 62
S₂₀ = 20/2 (5 + 62) = 670

Example 2: Same AS: find the least n such that Sₙ > 1000.

n/2 [10 + 3(n − 1)] > 1000
3n² + 7n − 2000 > 0
roots: n = [−7 ± √(49 + 24000)]/6,  positive root ≈ 24.7
least n = 25        check: S₂₄ = 948, S₂₅ = 1025 ✓

Example 3: GS: 12 + 6 + 3 + … Find the sum to infinity.

a = 12, r = 1/2, so S∞ = 12/(1 − ½) = 24.

Example 4: GS: 3, 6, 12, … Find the least n such that Tₙ > 10 000.

3(2ⁿ⁻¹) > 10 000
2ⁿ⁻¹ > 3333.3…
(n − 1) log 2 > log 3333.3…
n − 1 > 11.70
least n = 13        check: T₁₂ = 6144, T₁₃ = 12 288 ✓

Example 5: Find a and d: an AS has T₃ = 11 and T₈ = 26. Find a, d and T₂₀. Then, for a GS with T₂ = 6 and T₅ = 48, find a and r.

AS:  a + 2d = 11
     a + 7d = 26
     subtract: 5d = 15, d = 3;  a = 11 − 6 = 5
     T₂₀ = 5 + 19 × 3 = 62

GS:  ar  = 6
     ar⁴ = 48
     divide: r³ = 8, r = 2;  a = 6 ÷ 2 = 3

Common mistakes

Using n instead of n − 1 in Tₙ.
Using S∞ when |r| ≥ 1 (the sum doesn't settle down).
Dividing a log inequality by a negative log (0 < r < 1) without flipping the sign.
Taking the smaller root of the quadratic, or rounding 24.7 down instead of up.
Mixing up "the nth term" and "the sum of n terms".

How the DSE tests it

Section B: find a and d (or r) from two conditions; then "least n" with a sum or term; often an AS whose terms also form a GS, or a sequence built from logs (log of a GS is an AS).
Paper 2: general term, sum, and sum-to-infinity items.
Real life: salary rises, savings plans, depreciation over n years.

Questions you might not ask in class

How can infinitely many numbers add to a finite number?