Geometric Series - Past Edexcel Exam Questions
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1 Geometric Series - Past Edexcel Exam Questions 1. The second and fourth terms of a geometric series are 7.2 and respectively. The common ratio of the series is positive. For this series, find (a) the common ratio, [2] (b) the first term, [2] (c) the sum of the first 50 terms, giving your answer to 3 decimal places, [2] (d) the difference between the sum to infinity and the sum of first 50 terms, giving your answer to 3 decimal places. [2] Question 6 - Jan (a) A geometric series has first term a and common ratio r. Prove that the sum of the first n terms of the series is a (1 r n ) 1 r Mr King will be paid a salary of 35,000 in the year Mr King s contract promises a 4% increase in salary every year, the first increase being given in 2006, so that his annual salaries form a geometric sequence. (b) Find, to the nearest 100, Mr King s salary in the year [2] Mr King will receive a salary each year from 2005 until he retires at the end of (c) Find, to the nearest 1000, the total amount of salary he will receive in the period from 2005 until he retires at the end of [4] [4] Question 9 - Jun The first term of a geometric series is 120. The sum to infinity of the series is 480. (a) Show that the common ratio, r, is 3 4. [3]
2 (b) Find, to 2 decimal places, the difference between the 5th and 6th terms. [2] (c) Calculate the sum of the first 7 terms. [2] The sum of the first n terms is greater than 300. (d) Calculate the smallest possible value of n. [4] Question 4 - Jan A geometric series has first term a and common ratio r. The second term of the series is 4 and the sum to infinity of the series is 25. (a) Show that 25r 2 25r + 4 = 0. [4] (b) Find the two possible values of r. [2] (c) Find the corresponding two possible values of a. [2] (d) Show that the sum, S n, of the first n terms of the series is given by S n = 25 (1 r n ) Given that r takes the larger of its two possible values, (e) find the smallest value of n for which S n exceeds 24. [2] [1] Question 9 - May A geometric series is a + ar + ar (a) Prove that the sum of the first n terms of this series is given by (b) Find S n = a (1 rn ) 1 r 10 k=1 100 ( 2 k). [4] [3]
3 (c) Find the sum to infinity of the geometric series [3] (d) State the condition for an infinite geometric series with common ratio r to be convergent. [1] Question 10 - Jan A trading company made a profit of 50,000 in 2006 (Year 1). A model for future trading predicts that profits will increase year by year in a geometric sequence with common ratio r, r > 1. The model therefore predicts that in 2007 (Year 2) a profit of 50000r will be made. (a) Write down an expression for the predicted profit in Year n. [1] The model predicts that in Year n, the profit made will exceed 200,000. (b) Show that n > log [3] log r Using the model with r = 1.09, (c) find the year in which the profit made will exceed 200,000. [2] (d) find the total of the profits that will be made by the company over the 10 years from 2006 to 2015 inclusive, giving your answer to the nearest 10,000. [3] Question 8 - May The fourth term of a geometric series is 10 and the seventh term of the series is 80. For this series, find (a) the common ratio, [2] (b) the first term, [2] (c) the sum of the first 20 terms, giving your answer to the nearest whole number.. [2]
4 Question 2 - Jan A geometric series has first term 5 and common ratio 4 5. Calculate (a) the 20th term of the series, to 3 decimal places, [2] (b) the sum to infinity of the series. [2] Given that the sum to k terms of the series is greater than 24.95, (c) show that k > log log 0.8, [4] (d) find the smallest possible value of k. [1] Question 6 - Jun The first three terms of a geometric series are (k + 4), k and (2k 15), where k is a positive constant. (a) Show that k 2 7k 60 = 0. [4] (b) Hence show that k = 12. [2] (c) Find the common ratio of this series. [2] (d) Find the sum to infinity of this series. [2] Question 9 - Jan The third term of a geometric sequence is 324 and the sixth term is 96. (a) Show that the common ratio of the sequence is 2. 3 [2] (b) Find the first term of the sequence. [2] (c) Find the sum of the first 15 terms of the sequence. [3] (d) Find the sum to infinity of the sequence. [2]
5 Question 5 - Jun A car was purchased for 18,000 on 1st January. On 1st January each following year, the value of the car is 80% of its value on 1st January in the previous year. (a) Show that the value of the car exactly 3 years after it was purchased is [1] The value of the car falls below 1000 for the first time n years after it was purchased. (b) Find the value of n. [3] An insurance company has a scheme to cover the cost of maintenance of the car. The cost is 200 for the first year, and for every following year the cost increases by 12% so that for the third year the cost of the scheme is (c) Find the cost of the scheme for the 5th year, giving your answer to the nearest penny. [2] (d) Find the total cost of the insurance scheme for the first 15 years. [3] Question 6 - Jan The adult population of a town is 25,000 at the end of Year 1. A model predicts that the adult population of the town will increase by 3% each year, forming a geometric sequence. (a) Show that the predicted adult population at the end of Year 2 is 25,750. [1] (b) Write down the common ratio of the geometric sequence. [1] The model predicts that Year N will be the first year in which the adult population of the town exceeds 40,000. (c) Show that (N 1) log 1.03 > log 1.6 [3]
6 (d) Find the value of N. [2] At the end of each year, each member of the adult population of the town will give 1 to a charity fund. Assuming the population model, (e) find the total amount that will be given to the charity fund for the 10 years from the end of Year 1 to the end of Year 10, giving your answer to the nearest [3] Question 9 - Jun The second and fifth term of a geometric series are 750 and -6 respectively. Find (a) the common ratio, [3] (b) the first term of the series, [2] (c) the sum to infinity of the series. [2] Question 3 - Jan The second and third terms of a geometric series are 192 and 144 respectively. For this series, find (a) the common ratio, [2] (b) the first term, [2] (c) the sum to infinity, [2] (d) the smallest value of n for which the sum of the first n terms of the series exceeds [4] Question 6 - May A geometric series has first term a = 360 and common ratio r = 7 8. Giving your answers to 3 significant figures where appropriate, find
7 (a) the 20th term of the series, [2] (b) the sum of the first 20 terms of the series, [2] (c) the sum to infinity of the series. [2] Question 1 - Jan A geometric series is a + ar + ar (a) Prove that the sum of the first n terms of this series is given by S n = a (1 rn ) 1 r The third and fifth terms of a geometric series are 5.4 and respectively and all the terms in the series are positive. For this series find, (b) the common ratio, [2] (c) the first term, [2] (d) the sum to infinity. [3] [4] Question 9 - May A company predicts a yearly profit of 120,000 in the year The company predicts that the yearly profit will rise each year by 5%. The predicted yearly profit forms a geometric sequence with common ratio (a) Show that the predicted profit in the year 2016 is 138,915. [1] (b) Find the first year in which the yearly predicted profit exceeds 200,000. [5] (c) Find the total predicted profit for the years 2013 to 2023 inclusive, giving your answer to the nearest pound. [3] Question 3 - Jan 2013
8 18. The first three terms of a geometric series are 4p, (3p + 15) and (5p + 20) respectively, where p is a positive constant. (a) Show that 11p 2 10p 225 = 0. [4] (b) Hence show that p = 5. [2] (c) Find the common ratio of this series. [2] (d) Find the sum of the first ten terms of the series, giving your answer to the nearest integer. [3] Question 5 - Jun The first term of a geometric series is 20 and the common ratio is 7 8. The sum to infinity of the series is S (a) Find the value of S. [2] The sum to N terms of the series is S N. (b) Find, to 1 decimal place, the value of S 12. [2] (c) Find the smallest value of N, for which S S N < 0.5 [4]. Question 6 - Jun 2014
9 Solutions 1. (a) r = 9 10 (b) a = 8 (c) S 50 = (d) (a) - (b) 39,400 (c) S 20 = 1, 042, (a) - (b) 9.49 (c) S 7 = (d) n = 4 4. (a) - (b) r = 1 5, r = 4 5 (c) a = 20, a = 5 (d) - (e) n = (a) - (b) S 10 = 204, 600 (c) S = 5 4 (d) r < 1 6. (a) U n = 50000r n 1 (b) - (c) 2023 (d) 760, (a) r = 2 (b) a = 5 4 (c) S 20 = 1, 310, 719
10 8. (a) U 20 = (b) S = 25 (c) - (d) k = (a) - (b) - (c) r = 3 4 (d) S = (a) - (b) a = 729 (c) S 15 = (d) S = (a) - (b) n = 13 (c) (d) (a) - (b) r = 1.03 (c) - (d) N = 17 (e) S 10 = 287, (a) r = 1 5 (b) a = 3750 (c) S = (a) r = 3 4 (b) a = 256 (c) S = 1024 (d) n = (a) U 20 = 28.5
11 (b) S 20 = 2680 (c) S = (a) - (b) r = 3 5 (c) a = 15 (d) S = (a) - (b) 2024 (c) S 21 = 1, 704, (a) - (b) - (c) r = 3 2 (d) S = (a) S = 160 (b) S 12 = (c) n = 44
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