Generating Sequences

Generating Sequences

A sequence is an ordered list of numbers, each called a term. Sequences are defined by rules that allow every term to be calculated. Two fundamental types of rule are used at GCSE: the term-to-term rule (each term found from the previous one) and the position-to-term rule (the \(n\)th term expressed directly as a formula in \(n\)).

Term-to-Term Rules

A term-to-term rule states how to get the next term from the current one. It requires a starting value as well as the rule. Common examples:

  • Arithmetic: add (or subtract) a fixed value each time. E.g. start at 3, add 4 → 3, 7, 11, 15, …
  • Geometric: multiply by a fixed value each time. E.g. start at 2, multiply by 3 → 2, 6, 18, 54, …
  • Other: square and add, etc. E.g. \( u_{n+1} = u_n^2 - 1 \), \( u_1 = 2 \) → 2, 3, 8, 63, …

Position-to-Term Rules (\(n\)th Term)

The \(n\)th term formula gives the value of any term directly from its position \(n\) in the sequence, without calculating all previous terms. For an arithmetic sequence with first term \(a\) and common difference \(d\):

\[ T_n = a + (n-1)d \]

This is a linear expression in \(n\). The value of \(d\) is the gradient — it equals the constant difference between consecutive terms.

5 8 11 14 17 n=1 n=2 n=3 n=4 n=5 +3 +3 +3 +3 Tₙ = 3n + 2

Finding the \(n\)th Term of an Arithmetic Sequence

  1. Find the common difference \(d\) (subtract any term from the next).
  2. The \(n\)th term has \(d\) as the coefficient of \(n\): write \(T_n = dn + c\).
  3. Find \(c\) by substituting \(n=1\): \( c = T_1 - d \).

Example: sequence 5, 8, 11, 14, … → \(d=3\), so \(T_n=3n+c\). With \(n=1\): \(3(1)+c=5 Rightarrow c=2\). So \(T_n=3n+2\).

Worked Examples

The rule is "multiply by 2 and subtract 1". The first term is 3. Write the first 5 terms.
\[ 3, 5, 9, 17, 33 \]

\(3 imes 2-1=5\), \(5 imes 2-1=9\), \(9 imes 2-1=17\), \(17 imes 2-1=33\).

A sequence has \( u_1 = 10 \) and rule \( u_{n+1} = \frac{u_n}{2} + 1 \). Find the first 4 terms.
\[ 10, 6, 4, 3 \]
Find the \(n\)th term of the sequence 7, 11, 15, 19, …

\(d = 4\). So \(T_n = 4n + c\). At \(n=1\): \(4+c=7 Rightarrow c=3\).

\[ T_n = 4n + 3 \]
Find the \(n\)th term of 20, 17, 14, 11, …

\(d = -3\). \(T_n = -3n + c\). At \(n=1\): \(-3+c=20 Rightarrow c=23\).

\[ T_n = -3n + 23 \]
For the sequence \(T_n = 4n+3\): find the 20th term. Is 87 a term? Is 90 a term?
\[ T_{20} = 4(20)+3 = 83 \]

Is 87 a term? \( 4n+3=87 Rightarrow 4n=84 Rightarrow n=21 \). Integer ✓ — 87 is the 21st term.

Is 90 a term? \( 4n+3=90 Rightarrow 4n=87 Rightarrow n=21.75 \). Not an integer ✗ — 90 is not a term.

 Key Takeaways

  • Term-to-term rule: generates each term from the previous one. Requires a starting value.
  • Position-to-term (nth term): allows any term to be found directly. For arithmetic sequences: \(T_n = dn + c\) where \(d\) is the common difference.
  • To find \(c\): substitute \(n=1\) and set equal to the first term: \(c = T_1 - d\).
  • To check whether a value \(k\) is in the sequence: solve \(T_n = k\) — if \(n\) is a positive integer, \(k\) is a term; if not, it is not.
  • The common difference \(d\) is the constant amount added at each step. Negative \(d\) gives a decreasing sequence.