Unit 2 Exam Practice
Sequences and Series
These questions are written in the style of IB examination questions, and they are harder than the review exercises. The marks for each part are shown in square brackets on the right, for example [3]. Paper 1 means no calculator; Paper 2 means a graphic display calculator (GDC) is allowed. Try each question under exam conditions first, then open the markscheme.
| Mark | Meaning |
|---|---|
| M | Method mark: a correct method is used, even with a small slip |
| A | Accuracy mark: a correct answer or line of working (usually needs the M mark before it) |
| R | Reasoning mark: a clear reason or justification is given |
| N | Mark for a correct final answer with no working shown |
| AG | “Answer given” in the question, so no extra mark is awarded for writing it |
The number after the letter is the number of marks, so A2 is worth 2 marks. In the exam, give answers exactly or correct to 3 significant figures (s.f.) unless the question says otherwise.
SL questions
1 Paper 1
The second term of a geometric sequence is \(12\). The sum to infinity of the corresponding series is \(64\). Let \(r\) be the common ratio.
(a) Show that \(16r^{2} - 16r + 3 = 0\). [3]
(b) Find the two possible values of \(r\) and the corresponding values of the first term. [3]
(c) For the sequence with the smaller value of \(r\), find the sum of all the terms after the third term. [2]
Total 8 marks
(a)
\(u_1 r = 12\) and \(\dfrac{u_1}{1 - r} = 64\) M1
\(u_1 = \dfrac{12}{r}\), so \(\dfrac{12}{r(1 - r)} = 64\) A1
\(64r - 64r^{2} = 12 \;\Rightarrow\; 16r^{2} - 16r + 3 = 0\) A1 AG
[3 marks]
(b)
\((4r - 1)(4r - 3) = 0\) M1
\(r = \dfrac{1}{4}\) or \(r = \dfrac{3}{4}\) (both satisfy \(|r| < 1\)) A1
\(u_1 = 48\) or \(u_1 = 16\) respectively A1
[3 marks]
(c)
\(u_4 = 48 \times \left(\dfrac{1}{4}\right)^{3} = \dfrac{3}{4}\), and the sum is \(\dfrac{u_4}{1 - r}\) (or \(S_\infty - S_3 = 64 - 63\)) M1
\(= \dfrac{\frac{3}{4}}{\frac{3}{4}} = 1\) A1
[2 marks]
2 Paper 1
The sum of the first \(n\) terms of a sequence \(u_1, u_2, u_3, \ldots\) is given by \(S_n = 2n^{2} + 7n\), for \(n \in \mathbb{Z}^{+}\).
(a) Find \(u_2\). [2]
(b) Show that \(u_n = 4n + 5\), and hence explain why the sequence is arithmetic. [4]
(c) Find the value of \(\displaystyle\sum_{r=11}^{20} u_r\). [2]
Total 8 marks
(a)
\(u_2 = S_2 - S_1 = 22 - 9\) M1
\(u_2 = 13\) A1
[2 marks]
(b)
\(u_n = S_n - S_{n-1}\) M1
\(= 2n^{2} + 7n - \left(2(n - 1)^{2} + 7(n - 1)\right) = 2n^{2} + 7n - \left(2n^{2} + 3n - 5\right)\) A1
\(= 4n + 5\), and this also holds for \(n = 1\) because \(u_1 = S_1 = 9\) A1 AG
\(u_{n+1} - u_n = 4\), a constant, so the sequence is arithmetic R1
[4 marks]
(c)
\(\displaystyle\sum_{r=11}^{20} u_r = S_{20} - S_{10}\) M1
\(= 940 - 270 = 670\) A1
Accept \(\dfrac{10}{2}(49 + 85) = 670\).
[2 marks]
3 Paper 2
Mei invests $8000 in account A, which pays a nominal annual interest rate of \(3.5\%\), compounded monthly.
(a) Find the value of Mei’s investment after \(5\) years. Give your answer to the nearest dollar. [2]
Account B pays a nominal annual interest rate of \(r\%\), compounded quarterly. An investment of $8000 in account B has the same value after \(5\) years as the investment in account A.
(b) Find the value of \(r\). [3]
(c) Find the least number of complete years needed for the investment in account A to be worth more than $12 000. [2]
Total 7 marks
(a)
\(FV = 8000 \times \left(1 + \dfrac{3.5}{1200}\right)^{60}\) M1
\(= \$9528\) \((9527.54\ldots)\) A1
[2 marks]
(b)
\(8000 \times \left(1 + \dfrac{r}{400}\right)^{20} = 9527.54\ldots\) (for example, using a TVM solver on a GDC) M1
\(1 + \dfrac{r}{400} = \left(1 + \dfrac{3.5}{1200}\right)^{3}\) A1
\(r = 3.51\) \((3.5102\ldots)\) A1
[3 marks]
(c)
\(8000 \times \left(1 + \dfrac{3.5}{1200}\right)^{12n} > 12\,000\), so \(n > 11.6\) \((11.601\ldots)\) M1
\(12\) years A1
[2 marks]
See 2.5.
4 Paper 1
An arithmetic sequence has first term \(a\) and common difference \(d\), where \(d \neq 0\). The first, third and ninth terms of this arithmetic sequence are the first three terms of a geometric sequence.
(a) Show that \(a = d\). [4]
(b) Find the common ratio of the geometric sequence. [2]
The second term of the arithmetic sequence is \(6\).
(c) Find the sum of the first \(20\) terms of the arithmetic sequence. [3]
(d) Show that the sum of the first \(n\) terms of the geometric sequence is \(\dfrac{3}{2}\left(3^{n} - 1\right)\). [2]
(e) Find the least value of \(n\) for which the sum of the first \(n\) terms of the geometric sequence is greater than \(10\,000\). [3]
Total 14 marks
(a)
\(\dfrac{a + 2d}{a} = \dfrac{a + 8d}{a + 2d}\) (equal ratios) M1 A1
\(a^{2} + 4ad + 4d^{2} = a^{2} + 8ad \;\Rightarrow\; 4d^{2} = 4ad\) A1
\(4d(d - a) = 0\) and \(d \neq 0\), so \(a = d\) A1 AG
[4 marks]
(b)
the terms are \(a,\ a + 2a,\ a + 8a\), that is \(a,\ 3a,\ 9a\) M1
\(r = 3\) A1
[2 marks]
(c)
\(a + d = 2a = 6\), so \(a = d = 3\) A1
\(S_{20} = \dfrac{20}{2}\left(2 \times 3 + 19 \times 3\right)\) M1
\(= 630\) A1
[3 marks]
(d)
\(S_n = \dfrac{3\left(3^{n} - 1\right)}{3 - 1}\) M1
\(= \dfrac{3}{2}\left(3^{n} - 1\right)\) A1 AG
[2 marks]
(e)
\(\dfrac{3}{2}\left(3^{n} - 1\right) > 10\,000 \;\Rightarrow\; 3^{n} > \dfrac{20\,003}{3} = 6667.66\ldots\) M1
\(3^{8} = 6561\) and \(3^{9} = 19\,683\) A1
\(n = 9\) A1
[3 marks]
5 Paper 2
Amara buys a car for $32 000. The value of the car decreases by \(15\%\) each year.
(a) Find the value of the car after \(4\) years. [2]
(b) Find the least number of complete years after which the value of the car is less than $10 000. [3]
At the start of each year, Amara pays $2400 into a savings account. The account pays \(4\%\) interest per year, compounded annually. No other money is paid in or taken out.
(c) Show that the amount in the account at the end of the \(n\)th year is \(\$62\,400\left(1.04^{n} - 1\right)\). [3]
(d) Find the year at the end of which the amount in the account first exceeds the value of the car. [3]
Amara’s brother also pays $2400 at the start of each year into an account that pays \(i\%\) interest per year, compounded annually. At the end of the \(6\)th year his account holds $18 000.
(e) Find the value of \(i\). [3]
Total 14 marks
(a)
\(32\,000 \times 0.85^{4}\) M1
\(= \$16\,704.20\) A1
[2 marks]
(b)
\(32\,000 \times 0.85^{n} < 10\,000\) M1
\(n > \dfrac{\ln 0.3125}{\ln 0.85} = 7.157\ldots\) A1
\(8\) years A1
[3 marks]
(c)
\(2400(1.04) + 2400(1.04)^{2} + \cdots + 2400(1.04)^{n}\) (the first payment earns interest for \(n\) years) M1
geometric series with \(u_1 = 2496\) and ratio \(1.04\): \(S_n = \dfrac{2496\left(1.04^{n} - 1\right)}{1.04 - 1}\) A1
\(= 62\,400\left(1.04^{n} - 1\right)\) A1 AG
[3 marks]
(d)
\(62\,400\left(1.04^{n} - 1\right) > 32\,000 \times 0.85^{n}\) (for example, the intersection of the two graphs on a GDC) M1
\(n > 5.13\) \((5.1289\ldots)\) A1
at the end of the \(6\)th year A1
Accept a table of values: end of year 5, \(13\,519 < 14\,199\); end of year 6, \(16\,556 > 12\,069\).
[3 marks]
(e)
\(2400\left(1 + \dfrac{i}{100}\right) + 2400\left(1 + \dfrac{i}{100}\right)^{2} + \cdots + 2400\left(1 + \dfrac{i}{100}\right)^{6} = 18\,000\) (or a TVM solver on a GDC, payments at the start of each year) M1 A1
\(i = 6.41\) \((6.4119\ldots)\) A1
[3 marks]
HL questions
This unit has no AHL content, so the HL questions test the same syllabus content as SL, but they are longer and need more algebra, proof and reasoning.
1 Paper 1
Consider the infinite geometric series \[ 1 + \frac{2x}{x + 3} + \left(\frac{2x}{x + 3}\right)^{2} + \left(\frac{2x}{x + 3}\right)^{3} + \cdots, \qquad x \neq -3. \]
(a) Find the set of values of \(x\) for which the series converges. [4]
(b) Find the value of \(x\) for which the sum to infinity of the series is \(3\). [3]
Total 7 marks
(a)
\(\left|\dfrac{2x}{x + 3}\right| < 1\) M1
\(|2x| < |x + 3| \;\Rightarrow\; 4x^{2} < (x + 3)^{2}\) M1
\(3x^{2} - 6x - 9 < 0 \;\Rightarrow\; (x - 3)(x + 1) < 0\) A1
\(-1 < x < 3\) A1
Award M1 A1 A0 for \(x < 3\) only, found by solving \(\dfrac{2x}{x + 3} < 1\) alone.
[4 marks]
(b)
\(S_\infty = \dfrac{1}{1 - \frac{2x}{x + 3}} = \dfrac{x + 3}{3 - x}\) M1 A1
\(x + 3 = 9 - 3x \;\Rightarrow\; x = \dfrac{3}{2}\), which lies in \(-1 < x < 3\) A1
[3 marks]
See 2.4.
2 Paper 1
An arithmetic sequence has first term \(u_1\) and common difference \(d\). The sum of its first \(n\) terms is \(S_n\). It is given that \(S_{2n} = 4S_n\) for all \(n \in \mathbb{Z}^{+}\).
(a) Show that \(d = 2u_1\). [4]
(b) Hence show that \(S_n = n^{2}u_1\). [2]
(c) Given that \(u_1 \neq 0\), find the value of \(\dfrac{\displaystyle\sum_{r=n+1}^{3n} u_r}{S_n}\). [2]
Total 8 marks
(a)
\(S_{2n} = \dfrac{2n}{2}\big(2u_1 + (2n - 1)d\big)\) and \(4S_n = 4 \times \dfrac{n}{2}\big(2u_1 + (n - 1)d\big)\) M1 A1
dividing by \(n\): \(2u_1 + (2n - 1)d = 4u_1 + (2n - 2)d\) A1
\(d = 2u_1\) A1 AG
[4 marks]
(b)
\(S_n = \dfrac{n}{2}\big(2u_1 + (n - 1)(2u_1)\big)\) M1
\(= \dfrac{n}{2}\left(2nu_1\right) = n^{2}u_1\) A1 AG
[2 marks]
(c)
\(\displaystyle\sum_{r=n+1}^{3n} u_r = S_{3n} - S_n = 9n^{2}u_1 - n^{2}u_1 = 8n^{2}u_1\) M1
\(8\) A1
[2 marks]
3 Paper 1
A sequence is defined by \(u_r = \ln\left(\dfrac{8^{r}}{2}\right)\), for \(r \in \mathbb{Z}^{+}\).
(a) Show that \(u_r = (3r - 1)\ln 2\), and hence show that the sequence is arithmetic. [3]
(b) Show that \(\displaystyle\sum_{r=1}^{n} u_r = \frac{n(3n + 1)}{2}\ln 2\). [2]
(c) Find the value of \(n\) for which \(\displaystyle\sum_{r=1}^{n} u_r = \ln\left(2^{100}\right)\). [3]
Total 8 marks
(a)
\(u_r = \ln 8^{r} - \ln 2 = 3r\ln 2 - \ln 2\) M1
\(= (3r - 1)\ln 2\) A1 AG
\(u_{r+1} - u_r = 3\ln 2\), a constant, so the sequence is arithmetic R1
[3 marks]
(b)
\(u_1 = 2\ln 2\) and \(d = 3\ln 2\), so \(S_n = \dfrac{n}{2}\big(4\ln 2 + 3(n - 1)\ln 2\big)\) M1
\(= \dfrac{n(3n + 1)}{2}\ln 2\) A1 AG
[2 marks]
(c)
\(\dfrac{n(3n + 1)}{2}\ln 2 = 100\ln 2\) M1
\(3n^{2} + n - 200 = 0 \;\Rightarrow\; (3n + 25)(n - 8) = 0\) A1
\(n = 8\) (because \(n \in \mathbb{Z}^{+}\)) A1
[3 marks]
4 Paper 1
An infinite geometric series has first term \(a\) and common ratio \(r\), where \(a \neq 0\) and \(|r| < 1\). The sum to infinity of the series is \(12\).
A second series is formed by squaring each term of the first series: \[ a^{2} + a^{2}r^{2} + a^{2}r^{4} + \cdots \] The sum to infinity of the second series is \(48\).
(a) Explain why the second series is geometric and why it converges. [2]
(b) Show that \(a = 4(1 + r)\). [3]
(c) Find the value of \(a\) and the value of \(r\). [3]
(d) Show that the sum of the terms in the odd positions of the first series (\(u_1 + u_3 + u_5 + \cdots\)) is twice the sum of the terms in the even positions (\(u_2 + u_4 + u_6 + \cdots\)). [3]
(e) Let \(S_n\) be the sum of the first \(n\) terms of the first series. Find the least value of \(n\) for which \(12 - S_n < 0.001\). [4]
Total 15 marks
(a)
each term is the previous term multiplied by \(r^{2}\), so the series is geometric with common ratio \(r^{2}\) A1
\(|r| < 1\), so \(0 \leq r^{2} < 1\), and the series converges R1
[2 marks]
(b)
\(\dfrac{a}{1 - r} = 12\) and \(\dfrac{a^{2}}{1 - r^{2}} = 48\) M1
dividing: \(\dfrac{a^{2}}{(1 - r)(1 + r)} \times \dfrac{1 - r}{a} = \dfrac{a}{1 + r}\) A1
\(\dfrac{a}{1 + r} = \dfrac{48}{12} = 4\), so \(a = 4(1 + r)\) A1 AG
[3 marks]
(c)
\(12(1 - r) = 4(1 + r)\) M1
\(r = \dfrac{1}{2}\) A1
\(a = 6\) A1
[3 marks]
(d)
odd positions: \(a + ar^{2} + ar^{4} + \cdots = \dfrac{a}{1 - r^{2}}\) M1
\(= \dfrac{6}{\frac{3}{4}} = 8\) A1
even positions: \(12 - 8 = 4\), and \(8 = 2 \times 4\) A1
Accept a general argument: the even-position sum is \(\dfrac{ar}{1 - r^{2}}\), which is \(r = \dfrac{1}{2}\) times the odd-position sum.
[3 marks]
(e)
\(12 - S_n = \dfrac{ar^{n}}{1 - r} = 12\left(\dfrac{1}{2}\right)^{n}\) M1 A1
\(12\left(\dfrac{1}{2}\right)^{n} < 0.001 \;\Rightarrow\; 2^{n} > 12\,000\) A1
\(2^{13} = 8192\) and \(2^{14} = 16\,384\), so \(n = 14\) A1
[4 marks]
5 Paper 2
Kai borrows $25 000. Interest is charged at a nominal annual rate of \(6\%\), compounded monthly. At the end of each month, the interest for that month is added to the amount owed, and then Kai makes a repayment of \(\$P\). Let \(B_n\) be the amount owed, in dollars, immediately after the \(n\)th repayment.
(a) Write down an expression for \(B_1\) in terms of \(P\). [1]
(b) Show that \(B_n = 25\,000(1.005)^{n} - 200P\left(1.005^{n} - 1\right)\). [4]
(c) The loan is fully repaid after \(60\) repayments. Find the value of \(P\). [2]
(d) Find the total amount of interest that Kai pays. [2]
Instead, Kai decides to repay $600 at the end of each month, except that the final repayment is smaller: it is just enough to clear the loan.
(e) Find the number of repayments that Kai makes. [3]
(f) Find the amount of the final repayment. [3]
Total 15 marks
(a)
\(B_1 = 25\,000 \times 1.005 - P = 25\,125 - P\) A1
[1 mark]
(b)
\(B_2 = (25\,125 - P)(1.005) - P = 25\,000(1.005)^{2} - 1.005P - P\) M1
\(B_n = 25\,000(1.005)^{n} - P\left(1 + 1.005 + 1.005^{2} + \cdots + 1.005^{n-1}\right)\) A1
geometric series with first term \(1\), ratio \(1.005\) and \(n\) terms: \(\dfrac{1.005^{n} - 1}{1.005 - 1}\) M1
\(= 200\left(1.005^{n} - 1\right)\), so \(B_n = 25\,000(1.005)^{n} - 200P\left(1.005^{n} - 1\right)\) A1 AG
[4 marks]
(c)
\(B_{60} = 0 \;\Rightarrow\; P = \dfrac{25\,000(1.005)^{60}}{200\left(1.005^{60} - 1\right)}\) (or a TVM solver on a GDC) M1
\(P = 483.32\) \((483.320\ldots)\) A1
[2 marks]
(d)
\(60 \times 483.32 - 25\,000\) M1
\(= \$3999.20\) A1
Accept \(\$3999\) or \(\$4000\) (3 s.f.).
[2 marks]
(e)
\(25\,000(1.005)^{n} - 120\,000\left(1.005^{n} - 1\right) \leq 0\) (or a TVM solver on a GDC) M1
\(1.005^{n} \geq \dfrac{120\,000}{95\,000}\), so \(n \geq 46.8\) \((46.839\ldots)\) A1
\(47\) repayments A1
[3 marks]
(f)
\(B_{46} = 25\,000(1.005)^{46} - 120\,000\left(1.005^{46} - 1\right)\) M1
\(= 501.50\) \((501.502\ldots)\) A1
final repayment \(= 501.50 \times 1.005 = \$504.01\) A1
[3 marks]