TMUA: Algebra at GCSE level
GCSE algebra: formulae, identities and sequences.
- Section 1 Part 2 · M1-M7 GCSE-level knowledge (Part 2) · spec M4
- Papers 1 and 2
- 6 practice questions
- No calculator
What the specification covers
- Rearranging formulae, identities vs equations
- nth term of linear and quadratic sequences
Key ideas
- An identity is true for every value of the variable; an equation for particular values.
- Quadratic sequences have a constant second difference \(2a\), where \(an^2\) is the leading term.
- Find \(b\) and \(c\) in \(an^2+bn+c\) by matching the first two terms.
Common mistakes
- The second difference is \(2a\), not \(a\).
- When rearranging, do the same operation to every term on both sides.
Exam tip
Check a formula for the \(n\)th term against the fourth term, which you did not use to find it.
Worked example
Worked example
The first four terms of a quadratic sequence are \(6, 6, 4, 0\). What is term number 12?
- \(-32\)
- \(-108\)
- \(-106\)
- \(-104\)
- \(-84\)
Answer: D: \(-104\)
First differences \(0, -2, -4\); second difference \(-2\), so the \(n\)th term is \(-n^2+bn+c\). Matching the first two terms gives \(b=3\), \(c=4\). Term 12 is \(-1\times12^2+3\times12+4=-104\).
Practice questions
Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.
Question 1
The first four terms of a quadratic sequence are \(-4, -7, -12, -19\). What is term number 20?
- \(-403\)
- \(-405\)
- \(-400\)
- \(-364\)
- \(-131\)
Show the answer and solution
Answer: A: \(-403\)
First differences \(-3, -5, -7\); second difference \(-2\), so the \(n\)th term is \(-n^2+bn+c\). Matching the first two terms gives \(b=0\), \(c=-3\). Term 20 is \(-1\times20^2\times20-3=-403\).
Question 2
The first four terms of a quadratic sequence are \(6, 16, 32, 54\). What is term number 20?
- \(1104\)
- \(1220\)
- \(406\)
- \(1222\)
- \(1228\)
Show the answer and solution
Answer: D: \(1222\)
First differences \(10, 16, 22\); second difference \(6\), so the \(n\)th term is \(3n^2+bn+c\). Matching the first two terms gives \(b=1\), \(c=2\). Term 20 is \(3\times20^2+1\times20+2=1222\).
Question 3
The first four terms of a quadratic sequence are \(-3, -6, -11, -18\). What is term number 15?
- \(-227\)
- \(-229\)
- \(-95\)
- \(-225\)
- \(-198\)
Show the answer and solution
Answer: A: \(-227\)
First differences \(-3, -5, -7\); second difference \(-2\), so the \(n\)th term is \(-n^2+bn+c\). Matching the first two terms gives \(b=0\), \(c=-2\). Term 15 is \(-1\times15^2\times15-2=-227\).
Question 4
The first four terms of a quadratic sequence are \(1, -5, -13, -23\). What is term number 12?
- \(-149\)
- \(-103\)
- \(-177\)
- \(-180\)
- \(-175\)
Show the answer and solution
Answer: E: \(-175\)
First differences \(-6, -8, -10\); second difference \(-2\), so the \(n\)th term is \(-n^2+bn+c\). Matching the first two terms gives \(b=-3\), \(c=5\). Term 12 is \(-1\times12^2-3\times12+5=-175\).
Question 5
The first four terms of a quadratic sequence are \(5, 14, 29, 50\). What is term number 20?
- \(1085\)
- \(1200\)
- \(386\)
- \(1202\)
- \(1208\)
Show the answer and solution
Answer: D: \(1202\)
First differences \(9, 15, 21\); second difference \(6\), so the \(n\)th term is \(3n^2+bn+c\). Matching the first two terms gives \(b=0\), \(c=2\). Term 20 is \(3\times20^2\times20+2=1202\).
Keep going
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