Revision Library

revisionlibrary.co.uk

HIGHER

8.5 The Nth Term of a Quadratic Sequence

EDEXCEL 1MA1 · Calculator allowed · about 100 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

The nth term of a sequence is given by 2n² + 1.

(a)Work out the first three terms of the sequence.(2)

(b)Write down whether the sequence is increasing or decreasing.(1)

(Total for Question 1 is 3 marks)

The nth term of Sequence A is n² + 3n. The nth term of Sequence B is 2n + 9.

(Total for Question 2 is 2 marks)

The nth term of a sequence is n² - n + 5.

(a)Work out the 1st and 2nd terms of the sequence.(2)

(b)Work out the 20th term of the sequence.(2)

(Total for Question 3 is 4 marks)

Here are the first five terms of a quadratic sequence:
3, 8, 15, 24, 35

(a)Find an expression, in terms of n, for the nth term of the sequence.(3)

(b)Work out the 12th term of the sequence.(1)

(Total for Question 4 is 4 marks)

Here are the first four terms of a quadratic sequence:
7, 16, 29, 46

(Total for Question 5 is 3 marks)

Here are the first five terms of a quadratic sequence:
23, 20, 15, 8, -1

(Total for Question 6 is 3 marks)

Here are the first five terms of a quadratic sequence:
5, 12, 25, 44, 69

(Total for Question 7 is 3 marks)

A quadratic sequence has nth term n² + 4. Zara says, 'The number 68 is a term in this sequence.'

(Total for Question 8 is 3 marks)

Here are the first four terms of a sequence:
3n² - n gives the nth term.

(a)Work out the 2nd and 4th terms of the sequence.(2)

(b)Find the value of n for which the nth term of the sequence is 140.(3)

(Total for Question 9 is 5 marks)

A tile pattern is built so that Pattern 1 uses 4 tiles, Pattern 2 uses 11 tiles, Pattern 3 uses 22 tiles and Pattern 4 uses 37 tiles. The number of tiles in Pattern n follows a quadratic sequence.

(a)Show that the number of tiles in Pattern n is 2n² + n + 1.(3)

(b)Find the number of tiles used in Pattern 10.(2)

(c)A pattern in this sequence uses 301 tiles. Work out the pattern number.(3)

(Total for Question 10 is 8 marks)

The number of diagonals, D, in a polygon with n sides is given by the formula D = n(n-3)/2.

(a)Show that this formula can be written as D = 0.5n² - 1.5n.(2)

(b)Work out the number of diagonals in a polygon with 12 sides.(2)

(c)A polygon has 90 diagonals. Form and solve an equation to find the number of sides of the polygon.(4)

(Total for Question 11 is 8 marks)

Here are the first five terms of a quadratic sequence:
1.5, 4, 7.5, 12, 17.5

(Total for Question 12 is 3 marks)

Sequence A has nth term n² + 2. Sequence B has nth term 4n - 1.

(Total for Question 13 is 4 marks)

The number of squares, S, needed to build Shape n in a growing pattern is given by S = n² + n + 2.

(a)Deema builds Shape 6. Work out how many squares she needs.(2)

(b)Deema has 100 squares. Work out the largest shape number she can build without running out of squares.(3)

(Total for Question 14 is 5 marks)

The nth term of a quadratic sequence is given by an² + bn + c, where a, b and c are constants. The 2nd term of the sequence is 7, the 4th term is 25, and the 6th term is 59.

(Total for Question 15 is 5 marks)

A bakery in Leeds models its monthly profit, P, in thousands of pounds, in month n by P(n) = -2n² + 40n - 150, for 1 ≤ n ≤ 15, where n = 1 represents January 2020 and each increase of 1 in n represents one further year (so n=1 is 2020, n=2 is 2021, and so on).

(a)Work out the profit predicted for 2023.(2)

(b)Work out the maximum profit the model predicts, and state the year in which it occurs.(3)

(c)The bakery will close if the profit is negative for two or more consecutive years. Determine whether the model predicts the bakery will close within the given range of n.(4)

(Total for Question 16 is 9 marks)

A quadratic sequence has nth term n² - 7n + 15.

(a)Work out the 5th term of the sequence.(1)

(b)Find the smallest value in the sequence, and state the term number(s) at which it occurs.(3)

(Total for Question 17 is 4 marks)

Sequence P has nth term 2n² - 5n + 3. Sequence Q has nth term n² + 4n - 17.

(Total for Question 18 is 4 marks)

The nth term of a quadratic sequence is n² + n + 41.

(a)Work out the first four terms of the sequence.(2)

(b)Ben notices that 43, 47, 53 and 61 are all prime numbers and says, 'Every term in this sequence is a prime number.' Find a value of n that shows Ben is wrong.(2)

(Total for Question 19 is 4 marks)

A quadratic sequence has nth term n² - 6n + 12. Tomasz says, 'Some term in this sequence is equal to 2.'

(Total for Question 20 is 3 marks)

A quadratic sequence has nth term n² + n + 8. Priya says, 'Every term in this sequence is even.'

(Total for Question 21 is 3 marks)

Mark scheme · 8.5 The Nth Term of a Quadratic Sequence

Question 1

(a) B1 at least two of 3, 9, 19 correct
(a) B1 all of 3, 9, 19 correct (dep on previous B1)
(a) Answer: 3, 9, 19
(b) B1 increasing (cao)
(b) Answer: Increasing

Question 2

M1 n² + 3n + 2n + 9 seen (adding corresponding terms)
A1 n² + 5n + 9 oe, fully simplified
Answer: n² + 5n + 9

Question 3

(a) B1 1st term = 5
(a) B1 2nd term = 7
(a) Answer: 5, 7
(b) M1 substituting n=20 into n²-n+5
(b) A1 385 cao
(b) Answer: 385

Question 4

(a) M1 finding the second difference (2) and halving it to get the coefficient of n² (a=1)
(a) M1 comparing the sequence to n² (1, 4, 9, 16, 25) to find the linear remainder 2n
(a) A1 n² + 2n oe
(a) Answer: n² + 2n
(b) B1 168 (ft from candidate's expression in (a))
(b) Answer: 168

Question 5

M1 finding the second difference (4) and halving it to get a=2
M1 comparing the sequence to 2n² (2, 8, 18, 32) to find the linear remainder 3n+2
A1 2n² + 3n + 2 oe
Answer: 2n² + 3n + 2

Question 6

M1 finding the second difference (-2) and halving it to get a=-1
M1 comparing the sequence to -n² (-1, -4, -9, -16, -25) to find the constant remainder 24
A1 24 - n² oe
Answer: 24 - n²

Question 7

M1 finding the second difference (6) and halving it to get a=3
M1 comparing the sequence to 3n² (3, 12, 27, 48, 75) to find the linear remainder -2n+4
A1 3n² - 2n + 4 oe
Answer: 3n² - 2n + 4

Question 8

M1 forming the equation n² + 4 = 68, i.e. n² = 64
M1 solving to get n = 8 (or n = ±8)
A1 correct conclusion: since n=8 is a positive integer, Zara is correct (cao)
Answer: Zara is correct (n = 8)

Question 9

(a) B1 10
(a) B1 44
(a) Answer: 10, 44
(b) M1 forming the equation 3n² - n - 140 = 0
(b) M1 solving using the quadratic formula or factorisation
(b) A1 n = 7 cao (rejecting the non-integer negative solution)
(b) Answer: n = 7

Question 10

(a) M1 finding the second difference (4) of 4, 11, 22, 37 and halving it to get a=2
(a) M1 comparing the sequence to 2n² (2, 8, 18, 32) to find the linear remainder n+1
(a) A1 cso: 2n² + n + 1 shown to fit all four given values
(a) Answer: 2n² + n + 1
(b) M1 substituting n=10 into 2n²+n+1
(b) A1 211 cao
(b) Answer: 211
(c) M1 forming the equation 2n² + n - 300 = 0
(c) M1 solving using the quadratic formula (or factorisation)
(c) A1 12 cao (rejecting the negative solution)
(c) Answer: 12

Question 11

(a) M1 expanding n(n-3) = n² - 3n and dividing by 2
(a) A1 cso: 0.5n² - 1.5n
(a) Answer: 0.5n² - 1.5n
(b) M1 substituting n=12 into n(n-3)/2
(b) A1 54 cao
(b) Answer: 54
(c) M1 forming the equation n(n-3)/2 = 90, i.e. n² - 3n - 180 = 0
(c) M1 solving using factorisation or the quadratic formula
(c) A1 n = 15 cao
(c) A1 rejecting n = -12 since the number of sides must be a positive whole number
(c) Answer: 15 sides

Question 12

M1 finding the second difference (1) and halving it to get a=0.5
M1 comparing the sequence to 0.5n² (0.5, 2, 4.5, 8, 12.5) to find the linear remainder n
A1 0.5n² + n oe, e.g. (n²+2n)/2
Answer: 0.5n² + n

Question 13

M1 setting n² + 2 = 4n - 1
M1 rearranging to n² - 4n + 3 = 0
M1 factorising (n-1)(n-3) = 0, or using the quadratic formula
A1 both n = 1 and n = 3 cao
Answer: n = 1 and n = 3

Question 14

(a) M1 substituting n=6 into n²+n+2
(a) A1 44 cao
(a) Answer: 44
(b) M1 forming the equation or inequality n² + n + 2 = 100 (or ≤ 100)
(b) M1 solving using the quadratic formula to obtain n = 9.41 (2 dp) or equivalent
(b) A1 n = 9, with justification that Shape 10 needs 112 squares which is too many
(b) Answer: Shape 9

Question 15

M1 forming three equations: 4a+2b+c=7, 16a+4b+c=25, 36a+6b+c=59
M1 eliminating c to obtain two equations in a and b (e.g. 6a+b=9 and 10a+b=17)
M1 solving simultaneously to find a = 2
A1 b = -3
A1 c = 5
Answer: a = 2, b = -3, c = 5

Question 16

(a) M1 substituting n=4 into -2n²+40n-150
(a) A1 -22 cao, i.e. a loss of GBP22,000
(a) Answer: -22 (a loss of GBP22,000)
(b) M1 finding n=10 using n=-b/(2a) or completing the square
(b) M1 substituting n=10 to find P(10)=50
(b) A1 correct link of n=10 to the year 2029
(b) Answer: Maximum profit GBP50,000 in 2029
(c) M1 setting up -2n²+40n-150 < 0 and simplifying to n²-20n+75 > 0
(c) M1 solving n²-20n+75=0 to find the critical values n=5 and n=15
(c) M1 correctly identifying that profit is negative for n=1,2,3,4 (i.e. n<5) within the given range
(c) A1 correct conclusion: yes, the model predicts closure, since 2020-2023 (4 consecutive years) all have negative profit
(c) Answer: Yes, the bakery is predicted to close

Question 17

(a) B1 5 cao
(a) Answer: 5
(b) M1 finding the vertex of the quadratic at n=3.5 using n=-b/(2a) (or an equivalent method)
(b) M1 evaluating the sequence at n=3 and n=4 (both give 3)
(b) A1 smallest value is 3, occurring at both n=3 and n=4
(b) Answer: Smallest value is 3, at n=3 and n=4

Question 18

M1 setting up 2n²-5n+3 = (n²+4n-17) + 20
M1 rearranging to n² - 9n = 0
M1 factorising n(n-9) = 0
A1 n = 9 cao (rejecting n=0 as not a valid term number)
Answer: n = 9

Question 19

(a) B1 at least two of 43, 47, 53, 61 correct
(a) B1 all of 43, 47, 53, 61 correct
(a) Answer: 43, 47, 53, 61
(b) M1 testing a suitable value of n, such as n=40, in n²+n+41
(b) A1 correct value shown to be non-prime, e.g. n=40 gives 1681 = 41 x 41, which is not prime
(b) Answer: n = 40 gives 1681 = 41 x 41, which is not prime

Question 20

M1 forming the equation n² - 6n + 12 = 2, i.e. n² - 6n + 10 = 0
M1 calculating the discriminant: b² - 4ac = 36 - 40 = -4
A1 cso: since the discriminant is negative there are no real solutions for n, so Tomasz is wrong
Answer: Tomasz is wrong; no real value of n satisfies the equation

Question 21

M1 factorising n² + n = n(n+1)
M1 stating that n and n+1 are consecutive integers, so one of them must be even, meaning n(n+1) is always even
A1 cso: concluding that n(n+1) + 8 is the sum of two even numbers, so it is always even, proving Priya is correct
Answer: Priya is correct; every term is even

The Nth Term of a Quadratic Sequence - Worksheets, Questions and Revision