Sequences and Proportion: Revision Notes

Condensed for the final weeks. For worked examples, use the Sequences and Proportion study guide.

Identifying a sequence

Constant FIRST difference   -> LINEAR       (nth term = dn + c)
Constant SECOND difference  -> QUADRATIC    (nth term = an^2 + bn + c)
Constant RATIO              -> GEOMETRIC    (nth term = ar^(n-1))

Always write the differences underneath the sequence before deciding. If neither the first nor the second difference is constant, check whether consecutive terms share a constant ratio instead, since that points to an exponential sequence rather than a polynomial one.

Linear sequences

nth term = (first difference) x n  +  (adjustment)

Sequence:  5, 8, 11, 14
Difference = 3, so start with 3n:  3, 6, 9, 12
Adjustment: each term is 2 more   ->  nth term = 3n + 2

Check with n = 1 and n = 4 before committing.

Quadratic sequences

Second difference = 2a, so a = second difference ÷ 2.

Sequence:      3,  8, 15, 24
1st diff:        5,  7,  9
2nd diff:          2,  2      ->  a = 1,  so start with n^2

n^2:           1,  4,  9, 16
Subtract:      2,  4,  6,  8   ->  this is 2n

nth term = n^2 + 2n

Exponential sequences

Constant ratio between consecutive terms means the sequence is exponential: nth term = ar^(n-1).

Sequence:  2, 6, 18, 54
Each term is 3x the previous  ->  ratio r = 3
nth term = a x 3^(n-1),  and a = first term = 2
nth term = 2 x 3^(n-1)

Special sequences to recognise

Sequence	Terms
Square numbers	1, 4, 9, 16, 25
Cube numbers	1, 8, 27, 64
Triangular numbers	1, 3, 6, 10, 15
Fibonacci	1, 1, 2, 3, 5, 8 (add the previous two)
Powers of 2	2, 4, 8, 16, 32

4024 also expects simple combinations of these types — for example, a sequence whose nth term is a linear term plus a quadratic term, such as Tₙ = n² + 3n. Substituting n = 1, 2, 3 gives the terms 4, 10, 18 — check any combined formula this way before relying on it in an answer.

Proportion

	Direct	Inverse
Statement	y ∝ x	y ∝ 1/x
Equation	y = kx	y = k/x
As x doubles	y doubles	y halves
Graph	Straight line through origin	Curve (hyperbola)

Method every time:

1. Write the equation with k.
2. Substitute the given pair to find k.
3. Rewrite the full equation.
4. Use it to answer the question.

Variations: y ∝ x² → y = kx²; y ∝ x³ → y = kx³; y ∝ √x → y = k√x; y ∝ 1/√x → y = k/√x. Same method every time — only the power on x changes.

Worked: y is directly proportional to x². When x = 3, y = 45.

y = kx^2
45 = k x 3^2 = 9k   ->  k = 5
y = 5x^2

When x = 5:  y = 5 x 5^2 = 5 x 25 = 125

Worked: y is inversely proportional to x. When x = 4, y = 3.

y = k/x
3 = k/4   ->  k = 12
y = 12/x

When x = 6:  y = 12/6 = 2

Exam traps

• Check both difference rows before assuming linear.
• For a quadratic, a is half the second difference.
• In inverse proportion, y = k/x — not y = kx with a negative.
• Always find k explicitly; skipping it is where errors creep in.
• “Proportional to the square” means x², not 2x.
• Confusing exponential sequences (constant ratio) with linear or quadratic ones (constant differences) — check the ratio between terms if the differences themselves are not constant.
• Using x when the question specifies x², √x, x³ or ³√x — always check exactly which power or root is stated before setting up the equation.

Related: Sequences and Proportion practice questions for further worked problems in this style.

Self-test

1. Find the nth term of 7, 12, 17, 22.
2. Find the nth term of 2, 6, 12, 20.
3. y ∝ x and y = 15 when x = 3. Find y when x = 8.
4. p is inversely proportional to q. p = 6 when q = 2. Find p when q = 4.
5. Name the sequence 1, 3, 6, 10, 15.
6. Find the nth term of the sequence 2, 6, 18, 54.
7. y is directly proportional to x². y = 45 when x = 3. Find y when x = 5.

Answers: 1. First difference 5 → 5n gives 5, 10, 15, 20; each term is 2 more → 5n + 2. 2. First differences 4, 6, 8; second difference 2 → a = 1, n² gives 1, 4, 9, 16; subtracting leaves 1, 2, 3, 4 = n → n² + n. 3. k = 5, so y = 5x → y = 40. 4. k = 12, p = 12/q → p = 3. 5. Triangular numbers. 6. Constant ratio 3, so exponential: 2 × 3ⁿ⁻¹. 7. k = 5, so y = 5x² → y = 125.