Arithmetic and Geometric Sequences and Series - Worksheets, Questions and Revision

14 original exam-style questions - 4 pages of questions with a full mark scheme - free printable PDF.

Download PDF Jump to mark scheme (page 5)Arithmetic and Geometric Sequences and Series revision guide

This topic is chapter 5 of IGCSE Maths Practice Book 1.

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2.4 Arithmetic and Geometric Sequences and Series

EDEXCEL 4MA1 · Calculator allowed · about 50 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

An arithmetic sequence used to model temperatures has nth term a_n = 12 - 4n. Find the first three terms of this sequence, giving each term explicitly.

(Total for Question 1 is 1 mark)

In a savings plan the monthly deposit follows an arithmetic sequence with first term a = 7 and common difference d = 5. Find the nth term of this arithmetic sequence, written in terms of n.

(Total for Question 2 is 1 mark)

A shop reduces the price of an item by £6 each week. The price is £90 at week 1. Treating the weekly prices as an arithmetic sequence, find the price at week 10 and state the units.

(Total for Question 3 is 2 marks)

An arithmetic sequence models weekly savings: first term £20, common difference -£2 (weekly withdrawal). Calculate the sum of the first 10 terms and give the units.

(Total for Question 4 is 2 marks)

Calculate the sum of the first 8 terms of the arithmetic sequence 3, 7, 11, ... using the arithmetic series formula. State your answer as an exact integer.

(Total for Question 5 is 2 marks)

A geometric sequence used to model population has first term a = 6 and common ratio r = 1/3. Find the general nth term a_n = ar^n-1 and evaluate a₄, giving an exact fractional answer.

(Total for Question 6 is 2 marks)

Find the least value of n such that the partial sum of the arithmetic series with first term a = 5 and common difference d = 3 exceeds 500. Use S_n = n/2(2a + (n - 1)d) and show your working.

(Total for Question 7 is 3 marks)

Calculate the sum of the first 6 terms of the geometric sequence with first term a = 8 and common ratio r = 1/2, using the finite geometric series formula S_n = a(1 - rⁿ)/(1 - r). Give your answer as a fraction and a decimal.

(Total for Question 8 is 3 marks)

A convergent geometric series models a payment stream with first term a = 5 and common ratio r = 0.6. Find the sum to infinity of this series, using S_infty = a/(1 - r). Give your answer with one decimal place.

(Total for Question 9 is 2 marks)

A population is modelled by a geometric decay where the initial population is 2500 and it decreases by 12% each year. Find the population after 5 years, giving your answer to the nearest whole number and using ar^n-1 with a as the first-year population.

(Total for Question 10 is 2 marks)

A convergent geometric series has sum to infinity 75 and first term 5. Find the common ratio r of this series, giving r as an exact fraction and as a decimal to 4 decimal places.

(Total for Question 11 is 3 marks)

For the geometric sequence with first term a = 3 and common ratio r = 4/5, calculate the sum of the first 6 terms. Give your answer exactly as a fraction and as a decimal to 3 decimal places.

(Total for Question 12 is 3 marks)

A geometric series has first term a = 6 and common ratio r = 1/2. Find the integer value of n for which the sum of the first n terms equals 11.25. Show your algebraic steps.

(Total for Question 13 is 3 marks)

Show that the geometric series with first term a = 2 and common ratio r = -1/3 converges, and find its sum to infinity. Give the exact value and a decimal to two decimal places.

(Total for Question 14 is 4 marks)

Mark scheme · 2.4 Arithmetic and Geometric Sequences and Series

Question 1

B1 terms: 8, 4, 0 cao
Answer: 8, 4, 0

Question 2

B1 nth term = 7 + (n - 1)5 = 5n + 2, cao
Answer: 5n + 2

Question 3

M1 uses a_n = a + (n - 1)d with a = 90, d = -6 and n = 10
A1 a₁₀ = 90 + 9(-6) = £36 cao
Answer: £36

Question 4

M1 uses S₁₀ = 10/2(2*20 + 9*(-2))
A1 S₁₀ = £110 cao
Answer: £110

Question 5

M1 uses S_n = n/2(2a + (n - 1)d) with n = 8, a = 3, d = 4
A1 S₈ = 8/2(2*3 + 7*4) = 4(6 + 28) = 136 cao
Answer: 136

Question 6

M1 writes a_n = 6*(1/3)^{n - 1} and substitutes n = 4
A1 a₄ = 6*(1/3)³ = 6*(1/27) = 2/9 cao
Answer: a_n = 6*(1/3)^n-1, a₄ = 2/9

Question 7

M1 writes inequality n/2(2*5 + (n - 1)3) > 500 and expands to a quadratic, e.g. (3/2)n² + (7/2)n - 500 > 0, oe
M1 shows testing or solving leads to n = 17 giving S₁₇ = 493 ≤ 500 and n = 18 giving S₁₈ = 549 > 500
A1 least integer n = 18 cao
Answer: n = 18

Question 8

M1 uses S₆ = 8(1 - (1/2)⁶)/(1 - 1/2)
M1 evaluates numerator 1 - 1/64 = 63/64 and divides by 1/2 giving 8*(63/64)*2
A1 S₆ = 63/4 = 15.75 cao
Answer: 63/4 or 15.75

Question 9

M1 uses S_infty = 5/(1 - 0.6)
A1 S_infty = 5/0.4 = 12.5 cao
Answer: 12.5

Question 10

M1 uses a₅ = 2500*(0.88)⁵ with r = 0.88 and n = 5
A1 population approximately 1319 (to nearest whole number) cao
Answer: Approximately 1319 people

Question 11

M1 uses formula 75 = 5/(1 - r) and rearranges to 1 - r = 5/75 = 1/15
M1 finds r = 1 - 1/15 = 14/15
A1 r = 14/15, approximately 0.9333 cao
Answer: r = 14/15, approximately 0.9333

Question 12

M1 uses S₆ = 3(1 - (4/5)⁶)/(1 - 4/5)
M1 evaluates (4/5)⁶ = 4096/15625 and forms S₆ = 3*(11529/15625)/(1/5) = 3*(11529/15625)*5
A1 S₆ = 34587/3125, approximately 11.068 cao
Answer: 34587/3125, approximately 11.068

Question 13

M1 uses S_n = 6(1 - (1/2)ⁿ)/(1 - 1/2) = 12(1 - (1/2)ⁿ) and sets equal to 11.25
M1 solves 12(1 - (1/2)ⁿ) = 11.25 to get 1 - (1/2)ⁿ = 0.9375 and hence (1/2)ⁿ = 0.0625
A1 n = 4 cao
Answer: n = 4

Question 14

M1 states condition for convergence |r| < 1 and shows |-1/3| = 1/3 < 1
M1 uses S_infty = a/(1 - r) with a = 2 and r = -1/3
A1 exact sum S_infty = 2/(1 + 1/3) = 2/(4/3) = 3/2 cao
A1 decimal 1.50 cao
Answer: S_infty = 3/2 = 1.50

Arithmetic and Geometric Sequences and Series - Worksheets, Questions and Revision

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2.4 Arithmetic and Geometric Sequences and Series

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