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2c · E2.7–E2.8

Sequences and proportion

Topic 2 Algebra and graphs · Core and Extended · Papers 1–4

🎯What you need to be able to do

  • Continue a sequence or pattern and describe its term-to-term rule.
  • Find and use the \(n\)th term of linear sequences, and of simple quadratic and cubic sequences.
  • Find the \(n\)th term of any quadratic, cubic or exponential sequence, and simple combinations of these EXTENDED.
  • Express direct and inverse proportion algebraically (including squares, cubes and roots) and use it to find unknowns EXTENDED.

📚The mathematics

Linear sequences

If the terms go up by the same amount \(d\) each time, the \(n\)th term is \( dn + c \): the common difference times \(n\), plus whatever makes the first term right. For 5, 9, 13, 17, …: \( d = 4 \), and \( 4(1) + c = 5 \) gives \( c = 1 \), so the \(n\)th term is \( 4n + 1 \). To test whether a number is in the sequence, set the \(n\)th term equal to it: \(n\) must be a positive whole number.

Quadratic sequences

If the second differences are constant, the sequence is quadratic, and the \(n^{2}\) coefficient is half the second difference. Subtract that \(an^{2}\) from each term; what is left is linear.

A table of differences for the sequence 3, 8, 15, 24, 35. The first differences are 5, 7, 9, 11 and the second differences are all 2, so the sequence contains 1 n squared. Subtracting n squared (1, 4, 9, 16, 25) leaves 2, 4, 6, 8, 10, which is 2n, so the nth term is n squared plus 2n.
Second difference 2, so \(n^{2}\); the remainder 2, 4, 6, 8, 10 is \(2n\). The \(n\)th term is \( n^{2} + 2n \).

Cubic and exponential sequences EXTENDED

Recognise the building blocks: cubes 1, 8, 27, 64, … (\(n^{3}\)); powers such as 2, 4, 8, 16, … (\(2^{n}\)). A sequence that multiplies by \(r\) each time, starting at \(a\), has \(n\)th term \( a \times r^{n - 1} \). Many exam sequences are a building block plus a constant or a linear part, so compare the terms with \(n^{2}\), \(n^{3}\) or \(2^{n}\) and look at what is left over.

Proportion EXTENDED

\( y \propto x \): \( y = kx \)
\( y \propto x^{2} \): \( y = kx^{2} \)
\( y \propto \sqrt{x} \): \( y = k\sqrt{x} \)
\( y \propto \dfrac{1}{x} \): \( y = \dfrac{k}{x} \)
\( y \propto \dfrac{1}{x^{2}} \): \( y = \dfrac{k}{x^{2}} \)

The method is always the same: write the equation with \(k\), find \(k\) from the given pair of values, rewrite the equation with the value of \(k\), then use it. “Inversely” means divide by the quantity; “the square of” applies to the variable, not to \(k\).

Three graphs for positive x. y = 2x is a straight line through the origin; y = half x squared is a curve through the origin getting steeper; y = 4 over x is a curve that falls towards the x-axis without touching it.
Direct proportion passes through the origin; inverse proportion never meets either axis.

✏️Worked example (non-calculator)

(a) Find the \(n\)th term of 5, 9, 13, 17, …, and explain whether 150 is a term. [3] (b) EXTENDED Find the \(n\)th term of 3, 8, 15, 24, 35, … [2] (c) EXTENDED \(y\) is inversely proportional to the square of \(x\), and \( y = 12 \) when \( x = 2 \). Find \(y\) when \( x = 4 \), and the positive value of \(x\) when \( y = 0.75 \). [4]

(a) \( 4n + 1 \). If \( 4n + 1 = 150 \), then \( n = 37.25 \), which is not a whole number, so 150 is not a term.

(b) Second differences are 2 (see the diagram), so \( n^{2} \); the remainder is \( 2n \). The \(n\)th term is \( n^{2} + 2n \).

(c) \( y = \dfrac{k}{x^{2}} \), and \( 12 = \dfrac{k}{4} \) gives \( k = 48 \), so \( y = \dfrac{48}{x^{2}} \). When \( x = 4 \), \( y = 3 \). When \( y = 0.75 \): \( x^{2} = \dfrac{48}{0.75} = 64 \), so \( x = 8 \).

Check it. Doubling \(x\) from 2 to 4 should divide \(y\) by \( 2^{2} = 4 \): \( 12 \div 4 = 3 \) ✓. And for (b), \( n = 5 \) gives \( 25 + 10 = 35 \) ✓.
Using the second difference as the coefficient. Second difference 2 means \( 1n^{2} \), not \( 2n^{2} \) — the coefficient is half the second difference.

📝Practise

Questions in the style of the current papers. EXTENDED marks Extended-only content.

1. (Non-calculator.) Write down the next two terms of 2, 6, 12, 20, 30, … [2]
Differences 4, 6, 8, 10, so the next are +12 and +14: 42, 56.
2. (Non-calculator.) Find the \(n\)th term of 20, 17, 14, 11, … [2]
Difference \(-3\): \( -3n + c \), with \( -3 + c = 20 \Rightarrow c = 23 \). So \( 23 - 3n \).
3. (Non-calculator.) Find the \(n\)th term of 0, 3, 8, 15, 24, … [2]
Each term is one less than a square number: \( n^{2} - 1 \).
4. (Non-calculator.) EXTENDED Find the \(n\)th term of (a) 3, 10, 29, 66, … (b) 3, 6, 12, 24, … [3]
(a) Compare with the cubes 1, 8, 27, 64: each term is 2 more, so \( n^{3} + 2 \). (b) Multiply by 2 each time from 3: \( 3 \times 2^{n - 1} \).
5. (Non-calculator.) EXTENDED Find the \(n\)th term of 4, 11, 22, 37, 56, … [3]
First differences 7, 11, 15, 19; second differences 4, so \( 2n^{2} \). \( 2n^{2} \) is 2, 8, 18, 32, 50; the remainder 2, 3, 4, 5, 6 is \( n + 1 \). So \( 2n^{2} + n + 1 \).
6. (Calculator.) EXTENDED \(y\) is directly proportional to the square root of \(x\). When \( x = 4 \), \( y = 6 \). Find \(y\) when \( x = 49 \). [3]
\( y = k\sqrt{x} \); \( 6 = 2k \Rightarrow k = 3 \). So \( y = 3\sqrt{49} = 21 \).
7. (Calculator.) EXTENDED The distance \(d\) m an object falls is proportional to the square of the time \(t\) s. It falls 20 m in 2 s. How long does it take to fall 125 m? [3]
\( d = kt^{2} \); \( 20 = 4k \Rightarrow k = 5 \). \( 125 = 5t^{2} \Rightarrow t^{2} = 25 \Rightarrow t = 5 \) s.
8. (Non-calculator.) EXTENDED The brightness \(B\) of a lamp is inversely proportional to the square of the distance \(d\) from it. Describe what happens to \(B\) when \(d\) is tripled. [1]
\( B = \dfrac{k}{d^{2}} \); replacing \(d\) by \(3d\) divides \(B\) by \( 3^{2} = 9 \). The brightness becomes one ninth of what it was.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Corbettmaths — the \(n\)th term of quadratic sequences
  • Desmos — plot \( y = kx^{2} \) and \( y = k/x^{2} \) with a slider on \(k\)