Fibonacci, Quadratic and Geometric Sequences
Fibonacci, Quadratic and Geometric Sequences
Beyond arithmetic sequences (constant first difference), three further sequence types are assessed at GCSE: Fibonacci-type sequences (each term is the sum of the two preceding ones), quadratic sequences (constant second difference), and geometric sequences (constant ratio between consecutive terms).
Fibonacci-Type Sequences
In a Fibonacci-type sequence, each term is the sum of the two immediately before it: \( u_{n+1} = u_{n-1} + u_n \). The classic Fibonacci sequence is 1, 1, 2, 3, 5, 8, 13, 21, … Any two starting values can seed a Fibonacci-type sequence. These sequences grow approximately exponentially for large \(n\).
Quadratic Sequences
A quadratic sequence has a constant second difference (the differences of the differences are equal). The first differences form an arithmetic sequence, and the \(n\)th term is a quadratic expression in \(n\). Identifying constant second differences is the diagnostic tool.
Geometric Sequences
In a geometric sequence, each term is obtained by multiplying the previous term by a fixed value called the common ratio \(r\). The \(n\)th term is \( T_n = a cdot r^{n-1} \) where \(a\) is the first term. The ratio is found by dividing any term by the one before it: \( r = T_{n+1}/T_n \).
- \(r > 1\): geometric growth (terms increase rapidly).
- \(0 < r < 1\): geometric decay (terms decrease toward zero).
- \(r < 0\): terms alternate in sign.
Worked Examples
A Fibonacci-type sequence starts 2, 5. Write the first 7 terms.
\[ 2, 5, 7, 12, 19, 31, 50 \]Each term is the sum of the two before it: \(2+5=7\), \(5+7=12\), \(7+12=19\), etc.
Confirm that 5, 11, 21, 35, 53 is quadratic. Find the next term.
First differences: 6, 10, 14, 18. Second differences: 4, 4, 4 — constant. Quadratic confirmed.
Next first difference: 18+4=22. Next term: 53+22=75.
Find the common ratio and 5th term of the geometric sequence 3, 6, 12, 24, …
\[ r = 6/3 = 2 \qquad T_5 = 3 \times 2^4 = 48 \]Find the common ratio of the sequence 80, 20, 5, 1.25, … and the 6th term.
\[ r = 20/80 = \frac{1}{4} \qquad T_6 = 80 \times \left(\frac{1}{4}\right)^5 = 80 \times \frac{1}{1024} = \frac{80}{1024} = \frac{5}{64} \]Key Takeaways
- Fibonacci-type: each term = sum of previous two. Any two starting values work.
- Quadratic sequence: constant second differences. First differences form an arithmetic sequence.
- Geometric sequence: constant ratio \(r = T_{n+1}/T_n\). nth term: \(T_n = a cdot r^{n-1}\).
- Diagnostic: constant first difference → arithmetic; constant second difference → quadratic; constant ratio → geometric.
- To extend a quadratic sequence: continue the second differences, use them to extend the first differences, then extend the terms.