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 1 is 1 mark)
2
An arithmetic sequence used to model temperatures has nth term an = 12 - 4n. Find the first three terms of this sequence, giving each term explicitly.
(Total for Question 2 is 1 mark)
3
A shop reduces the price of an item by 6 pounds each week. The price is 90 pounds 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)
4
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 4 is 2 marks)
5
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 Sn = n/2(2a + (n - 1)d) and show your working.
(Total for Question 5 is 3 marks)
6
A geometric sequence used to model population has first term a = 6 and common ratio r = 1/3. Find the general nth term an = arn-1 and evaluate a4, giving an exact fractional answer.
(Total for Question 6 is 2 marks)
7
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 Sn = a(1 - rn)/(1 - r). Give your answer as a fraction and a decimal.
(Total for Question 7 is 3 marks)
8
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 8 is 2 marks)
9
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 arn-1 with a as the first-year population.
(Total for Question 9 is 2 marks)
10
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 10 is 3 marks)
11
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 11 is 3 marks)
12
An arithmetic sequence models weekly savings: first term 20 pounds, common difference -2 pounds (weekly withdrawal). Calculate the sum of the first 10 terms and give the units.
(Total for Question 12 is 2 marks)
13
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)
14
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 · IG.M9 Arithmetic and Geometric Sequences and Series
Question 1
B1 nth term = 7 + (n - 1)5 = 5n + 2, cao
Answer: 5n + 2
Question 2
B1 terms: 8, 4, 0 cao
Answer: 8, 4, 0
Question 3
M1 uses an = a + (n - 1)d with a = 90, d = -6 and n = 10
A1 a10 = 90 + 9(-6) = 36 pounds cao
Answer: 36 pounds
Question 4
M1 uses Sn = 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 5
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 6
M1 writes an = 6*(1/3)n - 1 and substitutes n = 4
A1 a4 = 6*(1/3)3 = 6*(1/27) = 2/9 cao
Answer: an = 6*(1/3)n-1, a4 = 2/9
Question 7
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 8
M1 uses S_infty = 5/(1 - 0.6)
A1 S_infty = 5/0.4 = 12.5 cao
Answer: 12.5
Question 9
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 10
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 11
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 12
M1 uses S10 = 10/2(2*20 + 9*(-2))
A1 S10 = 110 pounds cao
Answer: 110 pounds
Question 13
M1 uses Sn = 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