Sequences & Quadratic Inequalities

A sequence begins 2, 5, 10, 17, 26, ... Find its nth term.

• An² + 2
• Bn² + 1✓
• C2n + 1
• Dn² − 1

Answer: The first differences are 3, 5, 7, 9 and the second difference is constant at 2, so the n² coefficient is 2 ÷ 2 = 1. Subtracting n² (1, 4, 9, 16, 25) from the sequence leaves 1 every time, so the nth term is n² + 1. Adding 2 instead of 1 overshoots the constant term, treating the sequence as linear with 2n + 1 ignores the changing differences, and n² − 1 uses the wrong sign for the constant.

A quadratic sequence has nth term n² − 3n + 4. Find the 6th term.

• A22✓
• B16
• C28
• D18

Answer: Substituting n = 6 gives 6² − 3(6) + 4 = 36 − 18 + 4 = 22. Sixteen comes from subtracting the constant term instead of adding it, 28 comes from adding 3n rather than subtracting it, and 18 comes from a slip in combining the three terms.

The nth term of a sequence is 3n² + 2. Find the value of n for which the term equals 302.

• A20
• B10✓
• C15
• D30

Answer: Setting 3n² + 2 = 302 gives 3n² = 300, so n² = 100 and n = 10, taking the positive root since n is a term number. Twenty comes from doubling the correct value instead of taking the square root, and 30 and 15 both come from arithmetic slips when isolating n².

A sequence begins 3, 8, 15, 24, 35, ... Find its nth term.

• An + 2
• Bn² + n
• Cn² − n
• Dn² + 2n✓

Answer: The first differences are 5, 7, 9, 11 and the second difference is constant at 2, giving an n² coefficient of 1. Subtracting n² (1, 4, 9, 16, 25) from the sequence leaves 2, 4, 6, 8, 10, which is 2n, so the nth term is n² + 2n. n² + n gives only 2 when n = 1, not 3, n² − n uses the wrong sign on the linear term, and n + 2 wrongly treats the sequence as linear rather than quadratic.

Find the first three terms of the sequence with nth term 2n² − n.

• A1, 6, 14
• B1, 7, 17
• C2, 8, 18
• D1, 6, 15✓

Answer: Substituting n = 1, 2, 3 gives 2(1) − 1 = 1, 2(4) − 2 = 6, and 2(9) − 3 = 15. The other options each contain one term with an arithmetic slip, such as adding instead of subtracting n, or miscalculating 2n² for one value of n.

A quadratic sequence has nth term an² + bn + c. Its first three terms are 5, 12, 23. Find the value of c.

• A5
• B3
• C1
• D2✓

Answer: The first differences are 7 and 11, so the second difference is 4, giving 2a = 4 and a = 2. Using the first term, a + b + c = 5 with a = 2 and b = 1 (found from the first difference) gives c = 2. Five confuses c with the first term of the sequence itself, 1 confuses c with the value of b, and 3 comes from a slip when rearranging the equations.

A sequence begins 4, 7, 12, 19, 28, ... Find its nth term.

• An² − 3
• Bn² + 2
• Cn² + 3✓
• D2n + 2

Answer: The first differences are 3, 5, 7, 9 and the second difference is constant at 2, so the n² coefficient is 2 ÷ 2 = 1. Subtracting n² (1, 4, 9, 16, 25) from the sequence leaves 3 every time, so the nth term is n² + 3. Treating the sequence as linear with 2n + 2 ignores the changing differences, n² − 3 uses the wrong sign for the constant, and n² + 2 uses the wrong constant term.

A sequence begins 1, 6, 15, 28, 45, ... Find its nth term.

• An² − n
• B2n² − 1
• C2n² + n
• D2n² − n✓

Answer: The first differences are 5, 9, 13, 17 and the second difference is constant at 4, so the n² coefficient is 4 ÷ 2 = 2. Subtracting 2n² (2, 8, 18, 32, 50) from the sequence leaves −1, −2, −3, −4, −5, which is −n, so the nth term is 2n² − n. Adding n instead of subtracting it flips the sign on the linear term, using an n² coefficient of 1 instead of 2 ignores the constant second difference, and 2n² − 1 replaces the linear term with a constant that doesn't fit every value.

A quadratic sequence has nth term an² + bn + c. Its first three terms are 4, 9, 18. Find the value of b.

• A1
• B2
• C3
• D−1✓

Answer: The first differences are 5 and 9, so the second difference is 4, giving 2a = 4 and a = 2. Using the linear-term relationship 3a + b = (first difference) with the first difference of 5 gives 6 + b = 5, so b = −1. One reverses the correct sign, three confuses b with c, and two confuses b with a.

The first term of a geometric sequence is 8 and the 4th term is 1. Find the common ratio.

• A1/4
• B1/2✓
• C1/8
• D3/2

Answer: The 4th term equals the first term multiplied by r³, so 8r³ = 1, giving r³ = 1/8 and r = 1/2. 1/4 comes from taking a square root instead of a cube root, 3/2 comes from an arithmetic slip when solving r³ = 1/8, and 1/8 stops at r³ without taking the cube root.

A geometric sequence has first term 4 and common ratio 1/2. Find the sum of the first four terms.

• A7.5✓
• B6.25
• C8.5
• D6.5

Answer: The first four terms are 4, 2, 1 and 0.5, and adding them gives 4 + 2 + 1 + 0.5 = 7.5. 8.5 and 6.5 both come from arithmetic slips when adding the four terms, and 6.25 comes from stopping the addition after three terms and rounding.

Solve the inequality x² < 25, giving your answer as an inequality in x.

• A0 < x < 5
• B−5 < x < 5✓
• Cx > −5
• D0 ≤ x ≤ 5

Answer: The critical values are x = 5 and x = −5, and since x² < 25 describes a region below the parabola between its roots, the solution is −5 < x < 5. 0 < x < 5 and 0 ≤ x ≤ 5 both wrongly ignore all the negative solutions, and x > −5 keeps only one of the two boundary conditions instead of the full double inequality.

Solve x² − 9 ≥ 0, expressing the solution set.

• Ax ≤ 3 or x ≥ −3
• Bx ≤ −9 or x ≥ 9
• Cx < −3 or x > 3
• Dx ≤ −3 or x ≥ 3✓

Answer: The critical values are x = 3 and x = −3, and because the parabola x² − 9 opens upward, it is at or above zero outside these roots, giving x ≤ −3 or x ≥ 3. Using 9 as the boundary forgets to take the square root of the constant term, swapping to strict inequalities forgets that the roots themselves satisfy ≥ 0, and swapping the two boundary values the wrong way round attaches the sign to the wrong branch.

Solve the inequality x² − 5x + 6 < 0.

• A2 < x < 3✓
• B1 < x < 4
• C−3 < x < −2
• D3 < x < 2

Answer: Factorising gives (x − 2)(x − 3) < 0, and since the parabola opens upward, it is negative between its roots, so the solution is 2 < x < 3. −3 < x < −2 uses the roots with the wrong sign throughout, 3 < x < 2 writes the two correct roots in the wrong order, and 1 < x < 4 comes from factorising the expression incorrectly.

Solve x² + 2x − 8 ≤ 0, giving your answer as an inequality.

• A−4 ≤ x ≤ 2✓
• B−4 ≤ x ≤ 4
• C−2 ≤ x ≤ 4
• D0 ≤ x ≤ 2

Answer: Factorising gives (x + 4)(x − 2) ≤ 0, with critical values x = −4 and x = 2, and since the parabola opens upward, it is at or below zero between its roots, giving −4 ≤ x ≤ 2. −2 ≤ x ≤ 4 swaps the two critical values around, 0 ≤ x ≤ 2 wrongly ignores all the negative solutions, and −4 ≤ x ≤ 4 comes from factorising the expression incorrectly to give matching-sized roots.

Find the set of values of x for which x² − x − 12 > 0.

• Ax < 4 or x > 3
• B−3 < x < 4
• Cx < −3 or x > 4✓
• Dx < 3 or x > 4

Answer: Factorising gives (x + 3)(x − 4) > 0, with critical values x = −3 and x = 4, and since the parabola opens upward, it is positive outside these roots, giving x < −3 or x > 4. −3 < x < 4 describes the region between the roots where the expression is negative instead, and the other two options each attach the wrong sign to one of the two critical values.