(a)State whether Sequence A is arithmetic, geometric, or neither of these. Give a reason for your answer.(1)
(b)State whether Sequence B is arithmetic, geometric, or neither of these. Give a reason for your answer.(1)
(c)State whether Sequence C is arithmetic, geometric, or neither of these. Give a reason for your answer.(1)
(Total for Question 2 is 3 marks)
3
Here are the first four terms of a sequence. 50, 44, 38, 32
(a)Write down the next term of the sequence.(1)
(b)Find an expression, in terms of n, for the nth term of the sequence.(2)
(c)Explain whether 0 is a term of the sequence.(2)
(Total for Question 3 is 5 marks)
4
The nth term of a sequence is 6n - 4.
(a)Work out the first three terms of the sequence.(2)
(b)Work out the 20th term of the sequence.(1)
(Total for Question 4 is 3 marks)
5
Special sequences of numbers include square numbers and triangular numbers.
(a)Write down the first five square numbers.(1)
(b)Write down the first five triangular numbers.(1)
(c)The nth triangular number is given by the formula n(n + 1) / 2. Use this formula to find the 12th triangular number.(2)
(Total for Question 5 is 4 marks)
6
The nth term of a sequence is n2 + 3.
(a)Work out the 4th term of the sequence.(1)
(b)Work out the 10th term of the sequence.(1)
(c)Explain why 20 is not a term in the sequence.(2)
(Total for Question 6 is 4 marks)
7
Here are the first five terms of a sequence. 3, 4, 7, 11, 18
(a)Describe, in words, how each term of the sequence, after the first two, is found from the previous two terms.(1)
(b)Find the next two terms of the sequence.(2)
(Total for Question 7 is 3 marks)
8
A sequence of patterns is built from matchsticks. Pattern 1 is a single triangle. Each new pattern joins one more triangle onto the end of a strip, sharing one edge with the triangle before it. Pattern 1 (one triangle) uses 3 matchsticks. Pattern 2 (two triangles) uses 5 matchsticks. Pattern 3 (three triangles) uses 7 matchsticks.
(a)Write down the number of matchsticks needed for Pattern 4.(1)
(b)Find an expression, in terms of n, for the number of matchsticks in Pattern n.(2)
(c)Zainab claims that Pattern 20 uses exactly 39 matchsticks. Show that Zainab is wrong, and find the correct number of matchsticks in Pattern 20.(2)
(Total for Question 8 is 5 marks)
9
The nth term of a sequence is 7n - 3.
(a)Is 74 a term of the sequence? You must show working to justify your answer.(3)
(b)Is 101 a term of the sequence? You must show working to justify your answer.(2)
(Total for Question 9 is 5 marks)
10
Arjun is training for a marathon. In week 1, he runs 8 km. Each week after that, he runs 3 km more than the week before.
(a)Work out how far Arjun runs in week 6.(2)
(b)Find an expression, in terms of n, for the distance, in km, Arjun runs in week n.(2)
(c)Arjun wants to know the first week in which he will run at least 50 km. Find this week number.(3)
(Total for Question 10 is 7 marks)
11
The table shows the number of dots used to make a sequence of patterns. Pattern number (n): 1, 2, 3, 4, 5 Number of dots: 2, 7, 12, 17, ?
(a)Complete the table by finding the number of dots in Pattern 5.(1)
(b)Find an expression, in terms of n, for the number of dots in Pattern n.(2)
(c)Find the pattern number that has 92 dots.(2)
(Total for Question 11 is 5 marks)
12
In an arithmetic sequence, the 3rd term is 17 and the 7th term is 33.
(a)Find the common difference of the sequence.(2)
(b)Find the first term of the sequence.(2)
(c)Find an expression, in terms of n, for the nth term of the sequence.(1)
(Total for Question 12 is 5 marks)
13
Here are the first four terms of an arithmetic sequence. 15, 11.5, 8, 4.5
(a)Find the common difference of the sequence.(1)
(b)Find an expression, in terms of n, for the nth term of the sequence.(2)
(c)Find the first term in the sequence that is negative.(3)
(Total for Question 13 is 6 marks)
14
Here are the first four terms of a geometric sequence. 5, 15, 45, 135
(a)Find the common ratio of the sequence.(1)
(b)Find the 6th term of the sequence.(2)
(c)Explain whether 3000 is a term of the sequence.(2)
(Total for Question 14 is 5 marks)
15
Here are the first five terms of a quadratic sequence. 4, 7, 14, 25, 40
(a)Find the second difference of the sequence.(1)
(b)Find an expression, in terms of n, for the nth term of the sequence.(3)
(c)Use your expression to find the 10th term of the sequence.(1)
(Total for Question 15 is 5 marks)
16
Here are the first four terms of a quadratic sequence. 23, 18, 11, 2
(a)Show that the second difference of the sequence is -2.(1)
(b)Find an expression, in terms of n, for the nth term of the sequence.(3)
(c)Work out the 8th term of the sequence.(2)
(Total for Question 16 is 6 marks)
17
The nth term of a sequence is 6n - 1. Prove algebraically that the sum of any three consecutive terms of this sequence is always a multiple of 3.
(Total for Question 17 is 3 marks)
18
The nth term of a sequence is n2 + 3n. Prove algebraically that the difference between the (n + 1)th term and the nth term is always an even number.
(Total for Question 18 is 3 marks)
19
The nth term of a sequence is given by T(n) = 2n2 - 5n + 1.
(a)Show that solving T(n) = 43 leads to the equation 2n2 - 5n - 42 = 0.(1)
(b)Solve 2n2 - 5n - 42 = 0 to find the value of n for which T(n) = 43, explaining why the other solution to the equation is not valid.(4)
(c)Write down the term of the sequence that is equal to 43.(1)
(Total for Question 19 is 6 marks)
Mark scheme · 4.7D Sequences (Nth Term): Fluency and Exam Drill
Question 1
(a) B1 18 and 22 both correct oe
(a) Answer: 18, 22
(b) B1 start at 2 and add 4 each time oe
(b) Answer: Start at 2 and add 4 each time (oe).
Question 2
(a) B1 arithmetic, since the terms have a common difference of 4 oe
(a) Answer: Arithmetic (common difference of 4)
(b) B1 geometric, since the terms have a common ratio of 3 oe
(b) Answer: Geometric (common ratio of 3)
(c) B1 neither, since the differences (3, 5, 7) are not constant and the ratios are not constant oe
(c) Answer: Neither (it is a quadratic sequence)
Question 3
(a) B1 26 cao
(a) Answer: 26
(b) M1 common difference of -6 used correctly, e.g. -6n + c oe
(b) A1 56 - 6n oe cao
(b) Answer: 56 - 6n
(c) M1 56 - 6n = 0 oe, or 56 / 6 evaluated
(c) A1 correct conclusion: no, since n = 9.33... (28/3) is not a positive integer oe
(c) Answer: No, 0 is not a term (n would be 9.33..., not a whole number)
Question 4
(a) M1 at least one correct substitution, e.g. n=1 gives 2
(a) A1 2, 8, 14 all correct cao
(a) Answer: 2, 8, 14
(b) B1 116 cao
(b) Answer: 116
Question 5
(a) B1 1, 4, 9, 16, 25 all correct oe
(a) Answer: 1, 4, 9, 16, 25
(b) B1 1, 3, 6, 10, 15 all correct oe
(b) Answer: 1, 3, 6, 10, 15
(c) M1 12 x 13 / 2 oe
(c) A1 78 cao
(c) Answer: 78
Question 6
(a) B1 19 cao
(a) Answer: 19
(b) B1 103 cao
(b) Answer: 103
(c) M1 n2 + 3 = 20 rearranged to n2 = 17 oe
(c) A1 correct conclusion: 17 is not a square number, so there is no integer value of n oe
(c) Answer: 20 is not a term, since n2 = 17 has no integer solution (17 is not a square number)
Question 7
(a) B1 each term (from the third term onwards) is found by adding together the two previous terms oe
(a) Answer: Each term is the sum of the two terms before it (a Fibonacci-type sequence).
(b) M1 11 + 18 = 29 found
(b) A1 29 and 47 both correct cao
(b) Answer: 29, 47
Question 8
(a) B1 9 cao
(a) Answer: 9
(b) M1 common difference of 2 used correctly, e.g. 2n + c oe
(b) A1 2n + 1 oe cao
(b) Answer: 2n + 1
(c) M1 substitutes n = 20 into their expression from part (b), e.g. 2(20) + 1
(c) A1 41 (not 39), so Zainab is wrong ft cao
(c) Answer: 41 matchsticks (Zainab is wrong)
Question 9
(a) M1 7n - 3 = 74 oe, or 7n = 77
(a) A1 n = 11
(a) B1 correct conclusion: yes, since n = 11 is a positive integer, 74 is the 11th term ft
(a) Answer: Yes, 74 is a term (the 11th term)
(b) M1 7n - 3 = 101 oe, or 7n = 104
(b) A1 correct conclusion: no, since n = 14.86... (104/7) is not a whole number ft
(b) Answer: No, 101 is not a term
Question 10
(a) M1 8 + 5 x 3 oe, or lists terms up to week 6
(a) A1 23 km cao
(a) Answer: 23 km
(b) M1 common difference of 3 used correctly, e.g. 3n + c oe
(b) A1 3n + 5 oe cao
(b) Answer: 3n + 5
(c) M1 3n + 5 ≥ 50 oe (ft their expression)
(c) M1 3n ≥ 45 oe, leading to n ≥ 15
(c) A1 week 15 cao
(c) Answer: Week 15
Question 11
(a) B1 22 cao
(a) Answer: 22
(b) M1 common difference of 5 used correctly, e.g. 5n + c oe
(b) A1 5n - 3 oe cao
(b) Answer: 5n - 3
(c) M1 5n - 3 = 92 oe, or 5n = 95 (ft their expression)
(c) A1 Pattern 19 cao
(c) Answer: Pattern 19
Question 12
(a) M1 (33 - 17) / 4 oe, or forms two correct simultaneous equations
(a) A1 4 cao
(a) Answer: 4
(b) M1 a + 2(their d) = 17 oe
(b) A1 9 cao
(b) Answer: 9
(c) B1 4n + 5 oe cao (ft their a and d)
(c) Answer: 4n + 5
Question 13
(a) B1 -3.5 cao
(a) Answer: -3.5
(b) M1 common difference of -3.5 used correctly, e.g. -3.5n + c oe
(b) A1 18.5 - 3.5n oe cao
(b) Answer: 18.5 - 3.5n
(c) M1 18.5 - 3.5n < 0 oe (ft their expression)
(c) M1 n > 5.28... (37/7) found, leading to n = 6
(c) A1 6th term, -2.5 cao
(c) Answer: The 6th term, which is -2.5
Question 14
(a) B1 3 cao
(a) Answer: 3
(b) M1 5 x 35 oe, or lists terms up to the 6th
(b) A1 1215 cao
(b) Answer: 1215
(c) M1 identifies the two terms either side of 3000 (1215 and 3645), or sets up 5 x 3n-1 = 3000 oe
(c) A1 correct conclusion: no, 3000 is not a term, e.g. it lies strictly between 1215 and 3645 oe
(c) Answer: No, 3000 is not a term of the sequence
Question 15
(a) B1 4 cao, shown from first differences 3, 7, 11, 15
(a) Answer: 4
(b) M1 coefficient of n2 = half the second difference = 2 identified
(b) M1 forms and solves simultaneous equations (or equivalent) for the remaining terms
(b) A1 2n2 - 3n + 5 oe cao
(b) Answer: 2n2 - 3n + 5
(c) B1 175 cao ft their expression
(c) Answer: 175
Question 16
(a) B1 first differences -5, -7, -9 shown, giving a second difference of -2 cso
(a) Answer: -2 (shown)
(b) M1 coefficient of n2 = half the second difference = -1 identified
(b) M1 forms and solves simultaneous equations (or equivalent) for the remaining terms
(b) A1 -n2 - 2n + 26 oe cao
(b) Answer: -n2 - 2n + 26
(c) M1 substitutes n=8 into their expression, e.g. -(8)2 - 2(8) + 26
(c) A1 -54 cao ft
(c) Answer: -54
Question 17
M1 correct expressions for three consecutive terms, e.g. 6n - 1, 6n + 5, 6n + 11 oe
M1 correctly sums and simplifies to 18n + 15 oe
A1 factorises as 3(6n + 5) and concludes this is always a multiple of 3, since 6n + 5 is an integer cso
Answer: Proven: the sum is 3(6n + 5), which is always a multiple of 3.