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Cambridge IGCSE Mathematics — 0580 Extended

Topic 2.7: Algebra and graphs — Sequences

Continuing a sequence

The subscript \(T_n\) means the \(n\)th term. Position \(n\) starts at \(1\) unless the question says otherwise. The term-to-term rule tells you how to get from one term to the next; the position-to-term (nth-term) rule gives any term directly.

Find the next term of \(3, 7, 11, 15, \ldots\)

Sequence 3, 7, 11, 15 with term-to-term plus 4, next term 19
Term-to-term rule \(+4\). Next term \(19\).

Linear nth term

If first differences are constant, the sequence is linear: \(T_n = dn + b\), where \(d\) is that common difference.

Method

  1. Find the common difference \(d\).
  2. Write \(T_n = dn + b\). Substitute \(n = 1\) (the first term) to find \(b\).
  3. Check \(n = 2\) and \(n = 3\).

Find the \(n\)th term of \(3, 7, 11, 15, \ldots\)

Linear nth term of 3, 7, 11, 15 is 4n minus 1
Difference \(4\), first term \(3 = 4 - 1\), so \(T_n = 4n - 1\).

Paper 2 (non-calculator)

If the question asks for the \(n\)th term, write \(4n - 1\). “Add 4 each time” is a term-to-term rule and does not score as an nth-term formula.

Quadratic sequences

If the second differences are constant, the sequence is quadratic. For \(n^2\) the second difference is \(+2\), so a second difference of \(+2\) means the coefficient of \(n^2\) is \(1\). Compare the sequence with \(n^2\), then adjust.

Find the \(n\)th term of \(2, 5, 10, 17, \ldots\)

Difference table for 2, 5, 10, 17 with second difference plus 2 giving n squared plus 1
First differences \(+3, +5, +7\); second difference \(+2\). Compare with \(n^2\): \(T_n = n^2 + 1\).

Cubic and exponential

Cubes grow as \(1, 8, 27, 64, \ldots\) so \(T_n = n^3\). If each term is multiplied by the same constant, the sequence is exponential (geometric).

Find the \(n\)th term of \(1, 8, 27, 64\) and of \(3, 6, 12, 24\).

Cubic sequence n cubed and exponential sequence 3 times 2 to the n minus 1
\(1, 8, 27, 64 = n^3\). \(3, 6, 12, 24 = 3 \times 2^{n-1}\).

Combinations

Some sequences mix a power with a constant. Use differences to rule out linear or quadratic first, then test simple combinations such as \(n^2 + 1\) or \(2^n + 1\).

Which method?

First differences are constant

Linear: \(T_n = dn + b\).

Second differences are constant

Quadratic: compare with \(n^2\), then adjust.

Each term is multiplied by the same constant

Exponential: \(T_n = a \times r^{n-1}\).

Find the \(n\)th term of \(3, 5, 9, 17, \ldots\)

Sequence 3, 5, 9, 17 identified as 2 to the n plus 1
Differences \(+2, +4, +8\) point to powers of \(2\). \(T_n = 2^n + 1\).

Try this

Find the \(n\)th term of \(5, 8, 11, 14, \ldots\)

Show answer
Answer

  1. Common difference is \(3\), so \(T_n = 3n + b\)

    \[ T_1 = 3(1) + b = 5 \]

  2. Solve for \(b\)

    \[ b = 2 \]

  3. Final answer

    \[ T_n = 3n + 2 \]

Exam Traps

  • \(n^2 + n\) and \((n+1)^2\) look similar. Substitute \(n = 1\): \(n^2 + n = 2\) but \((n+1)^2 = 4\). Always check the first term.
  • Position \(n\) starts at \(1\) unless the question numbers from \(0\) or from a given \(n\).

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