4.7H Sequences (Nth Term): Higher Tier Practice

EDEXCEL Edexcel 1MA1 · Calculator allowed · about 75 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Work out the 6th term of the arithmetic sequence with nth term 7n - 4.

(Total for Question 1 is 2 marks)

The first three terms of a sequence are 3, 8, 15.
(a) Work out the 4th term.
(b) Suggest an nth term formula for this sequence.

(a)Work out the 4th term.(1)

(b)Suggest an nth term formula for this sequence.(2)

(Total for Question 2 is 3 marks)

The nth term of a sequence is 4n² - 2n.
(a) Work out the 1st and 2nd terms.
(b) Show that the difference between consecutive terms is 8n + 2 when going from term n to term n+1. Give a short explanation.

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

(b)Show that the difference between consecutive terms is 8n + 2 when going from term n to term n+1. Give a short explanation.(2)

(Total for Question 3 is 4 marks)

A sequence has nth term given by T_n = 5n - 3 for n odd, and T_n = 5n + 1 for n even.
(a) Work out T₅ and T₆.
(b) Explain briefly whether this sequence is arithmetic overall.

(a)Work out T₅ and T₆.(2)

(b)Explain briefly whether this sequence is arithmetic overall.(2)

(Total for Question 4 is 4 marks)

Find the nth term of the sequence: 11, 18, 25, 32, ...
Give your answer in the form an + b.

(Total for Question 5 is 5 marks)

A sequence is defined recursively by U₁ = 6 and U_{n+1} = 2U_n + 1 for n ≥ 1.
(a) Work out U₂ and U₃.
(b) Show that U_n = 7*2^{n-1} - 1 for all n ≥ 1. You may use proof by induction or an argument that checks the base case and shows the recurrence preserves the form.

(a)Work out U₂ and U₃.(2)

(b)Show that U_n = 7*2^{n-1} - 1 for all n ≥ 1. You may use proof by induction or an argument that checks the base case and shows the recurrence preserves the form.(4)

(Total for Question 6 is 6 marks)

Mark scheme · 4.7H Sequences (Nth Term): Higher Tier Practice

Question 1

M1 substitute n = 6 into 7n - 4 or show 7 x 6 - 4
A1 38 cao
Answer: 38

Question 2

(a) B1 24 cao
(a) Answer: 24
(b) M1 recognise quadratic form with second difference 2 and write an² + bn + c with a = 1
(b) A1 n² + 2n cao
(b) Answer: n² + 2n

Question 3

(a) M1 substitute n = 1 and n = 2 into 4n² - 2n
(a) A1 2 and 12 cao
(a) Answer: 1st term = 2, 2nd term = 12
(b) M1 compute T(n+1) - T(n) = [4(n+1)² - 2(n+1)] - [4n² - 2n]
(b) A1 8n + 2 cao with brief simplification shown
(b) Answer: 8n + 2

Question 4

(a) M1 substitute n = 5 and n = 6 into correct formula depending on parity
(a) A1 22 and 31 cao
(a) Answer: T₅ = 22, T₆ = 31
(b) M1 compare differences between consecutive terms and note they are not constant
(b) A1 statement: not arithmetic because differences are not constant (example given)
(b) Answer: Not arithmetic, because consecutive differences are not constant (they alternate).

Question 5

M1 identify common difference 7 and write general linear form an + b
M1 state a = 7 and substitute n = 1 into a + b = 11 to find b
M1 compute b = 11 - 7 = 4
A1 7n + 4 cao
B1 final answer stated clearly
Answer: 7n + 4

Question 6

(a) M1 substitute U₁ = 6 into recurrence to get U₂, then substitute U₂ to get U₃
(a) A1 13 and 27 cao
(a) Answer: U₂ = 13, U₃ = 27
(b) M1 base case: verify n=1 gives U₁ = 7*2^{0} - 1 = 7 - 1 = 6
(b) M1 assume for some k that U_k = 7*2^{k-1} - 1 (induction hypothesis)
(b) M1 use recurrence: U_{k+1} = 2U_k + 1 and substitute hypothesis to get 2(7*2^{k-1} - 1) + 1 = 7*2^{k} - 1
(b) A1 conclude U_{k+1} has the required form so formula holds for all n ≥ 1
(b) Answer: 7*2^{n-1} - 1

Sequences (Nth Term): Higher Tier Practice - Worksheets, Questions and Revision