Practice Questions on Sequences & Series
IB Math AA SL Style Questions with Step-by-Step Solutions
Below are ten practice questions, each similar in style to an IB Math AA SL problem on sequences and series. They focus on arithmetic or geometric sequences. Step-by-step solutions are provided, with relevant formulas clearly stated.
Question 1 [6 marks]
An arithmetic sequence has first three terms \(u_1 = 20\), \(u_2 = 17\), and \(u_3 = 14\).
- [2 marks] Find the common difference \(d\).
- [2 marks] Find \(u_6\).
- [2 marks] Find the sum of the first 6 terms, \(S_6\).
Solution:
- Common difference \(d\)
From an arithmetic sequence, \(d = u_2 - u_1 = 17 - 20 = -3\). - Find \(u_6\)
General term of an arithmetic sequence: \[ u_n = u_1 + (n-1)d. \] For \(n=6\): \[ u_6 = 20 + (6-1)(-3) = 20 + 5 \cdot (-3) = 20 - 15 = 5. \] - Find \(S_6\)
The sum of the first \(n\) terms in an arithmetic sequence is: \[ S_n = \frac{n}{2} \bigl(u_1 + u_n\bigr). \] So, \[ S_6 = \frac{6}{2}\bigl(u_1 + u_6\bigr) = 3 \times (20 + 5) = 3 \times 25 = 75. \]
Question 2 [6 marks]
An arithmetic sequence has first three terms \(u_1 = 7\), \(u_2 = 3\), and \(u_3 = -1\).
- [2 marks] Determine the common difference \(d\).
- [2 marks] Find \(u_7\).
- [2 marks] Calculate the sum of the first 7 terms, \(S_7\).
Solution:
- Common difference \(d\)
Since \(u_2 - u_1 = 3 - 7 = -4\), we have \(d = -4\). - Find \(u_7\)
Using \(u_n = u_1 + (n-1)d\): \[ u_7 = 7 + (7-1)(-4) = 7 + 6 \cdot (-4) = 7 - 24 = -17. \] - Find \(S_7\)
Recall: \[ S_n = \frac{n}{2}(u_1 + u_n). \] Here, \(n=7\): \[ S_7 = \frac{7}{2} \bigl(7 + (-17)\bigr) = \frac{7}{2} \times (-10) = 7 \times (-5) = -35. \]
Question 3 [6 marks]
A geometric sequence has first three terms \(u_1 = 4\), \(u_2 = 12\), and \(u_3 = 36\).
- [2 marks] Find the common ratio \(r\).
- [2 marks] Find \(u_5\).
- [2 marks] Find the sum of the first 5 terms, \(S_5\).
Solution:
- Common ratio \(r\)
In a geometric sequence, \(r = \frac{u_2}{u_1} = \frac{12}{4} = 3.\) - Find \(u_5\)
General term for a geometric sequence: \[ u_n = u_1 \, r^{\,n-1}. \] For \(n=5\): \[ u_5 = 4 \times 3^{\,5-1} = 4 \times 3^4 = 4 \times 81 = 324. \] - Find \(S_5\)
Sum of the first \(n\) terms in a geometric sequence (for \(r \neq 1\)): \[ S_n = u_1 \frac{r^n - 1}{r - 1}. \] Here \(n=5\), \(u_1=4\), and \(r=3\): \[ S_5 = 4 \times \frac{3^5 - 1}{3 - 1} = 4 \times \frac{243 - 1}{2} = 4 \times \frac{242}{2} = 4 \times 121 = 484. \]
Question 4 [6 marks]
A geometric sequence has \(u_1 = 16\), \(u_2 = 8\), and \(u_3 = 4\).
- [2 marks] Determine the common ratio \(r\).
- [2 marks] Find \(u_6\).
- [2 marks] Find the sum of the first 6 terms, \(S_6\).
Solution:
- Common ratio \(r\)
Compute \(r = \frac{u_2}{u_1} = \frac{8}{16} = \tfrac{1}{2}.\) - Find \(u_6\)
\[ u_n = u_1 \, r^{\,n-1}. \] For \(n=6\): \[ u_6 = 16 \left(\tfrac{1}{2}\right)^{5} = 16 \times \tfrac{1}{32} = \tfrac{16}{32} = \tfrac{1}{2}. \] - Find \(S_6\)
Using \[ S_n = u_1 \frac{r^n - 1}{r - 1} \quad (\text{for }r \neq 1), \] we get \[ S_6 = 16 \times \frac{\left(\tfrac{1}{2}\right)^6 - 1}{\tfrac{1}{2} - 1} = 16 \times \frac{\tfrac{1}{64} - 1}{- \tfrac{1}{2}} = 16 \times \frac{- \tfrac{63}{64}}{- \tfrac{1}{2}}. \] Simplify step by step: \[ -\tfrac{63}{64} \div -\tfrac{1}{2} = \tfrac{63}{64} \times 2 = \tfrac{63}{32}. \] So \[ S_6 = 16 \times \tfrac{63}{32} = \tfrac{16 \times 63}{32} = \tfrac{16}{32} \times 63 = \tfrac{1}{2} \times 63 = 31.5. \] Or written as fraction: \[ S_6 = 31.5 \quad \text{or} \quad \tfrac{63}{2}. \]
Question 5 [6 marks]
Consider an arithmetic sequence where the first term \(u_1 = 5\) and the common difference \(d = 2\).
- [2 marks] Write down the first 3 terms of the sequence.
- [2 marks] Find \(u_{10}\), the 10th term.
- [2 marks] Calculate the sum of the first 10 terms, \(S_{10}\).
Solution:
- First 3 terms
\(u_1 = 5.\)
\(u_2 = 5 + 2 = 7.\)
\(u_3 = 7 + 2 = 9.\) - Find \(u_{10}\)
General term: \(u_n = u_1 + (n-1)d.\)
For \(n=10\): \[ u_{10} = 5 + (10-1)\times 2 = 5 + 9\times 2 = 5 + 18 = 23. \] - Calculate \(S_{10}\)
Sum formula: \[ S_n = \frac{n}{2} (u_1 + u_n). \] Here \(n=10\): \[ S_{10} = \frac{10}{2} (5 + 23) = 5 \times 28 = 140. \]
Question 6 [Maximum mark: 6]
Consider an arithmetic sequence: 5, 8, 11, 14, …
- (a) Find the common difference, d.
- (b) Find the 10th term in the sequence.
- (c) Find the sum of the first 10 terms in the sequence.
Solution:
(a) We see that \(u_1 = 5\) and \(u_2 = 8\). The common difference is
\(d = u_2 - u_1 = 8 - 5 = 3\).
(b) The nth term of an arithmetic sequence is \[ u_n = u_1 + (n - 1)d. \] For \(n = 10\): \[ u_{10} = 5 + (10 - 1)\times 3 = 5 + 9\times 3 = 5 + 27 = 32. \]
(c) The sum of the first n terms is \[ S_n = \frac{n}{2}\bigl(u_1 + u_n\bigr). \] Thus, for \(n = 10\): \[ S_{10} = \frac{10}{2} \bigl(5 + 32\bigr) = 5 \times 37 = 185. \]
Question 7 [Maximum mark: 6]
Consider an arithmetic sequence: 0, -4, -8, -12, …
- (a) Find the common difference, d.
- (b) Find the 10th term in the sequence.
- (c) Find the sum of the first 10 terms in the sequence.
Solution:
(a) We observe
\(u_1 = 0\) and \(u_2 = -4\). Hence:
\(d = u_2 - u_1 = -4 - 0 = -4.\)
(b) Using
\(\displaystyle u_n = u_1 + (n-1)d\):
\(u_{10} = 0 + (10-1)\times (-4) = 9 \times (-4) = -36.\)
(c) The sum of the first n terms formula:
\(\displaystyle S_n = \frac{n}{2}\bigl(u_1 + u_n\bigr)\).
For \(n=10\):
\(\displaystyle S_{10} = \frac{10}{2}\,\bigl(0 + (-36)\bigr) = 5 \times (-36) = -180.\)
Question 8 [Maximum mark: 6]
Consider an arithmetic sequence: 12, 9, 6, 3, …
- (a) Find the common difference, d.
- (b) Find the 10th term in the sequence.
- (c) Find the sum of the first 10 terms in the sequence.
Solution:
(a)
Here \(u_1 = 12\) and \(u_2 = 9\). So
\(d = u_2 - u_1 = 9 - 12 = -3.\)
(b)
Using \(u_n = u_1 + (n - 1)d\):
\(u_{10} = 12 + (10 - 1)\times (-3) = 12 + 9\times(-3) = 12 - 27 = -15.\)
(c)
The sum formula: \(\displaystyle S_n = \frac{n}{2}(u_1 + u_n)\).
So for \(n=10\):
\(\displaystyle S_{10} = \frac{10}{2}(12 + (-15)) = 5 \times (-3) = -15.\)
Question 9 [Maximum mark: 6]
Consider an arithmetic sequence: 1, 3, 5, 7, …
- (a) Find the common difference, d.
- (b) Find the 10th term in the sequence.
- (c) Find the sum of the first 10 terms in the sequence.
Solution:
(a)
From \(u_1 = 1\) and \(u_2 = 3\), we get
\(d = 3 - 1 = 2.\)
(b)
Using \(u_n = u_1 + (n-1)d\):
\(u_{10} = 1 + (10 - 1)\times 2 = 1 + 9\times 2 = 1 + 18 = 19.\)
(c)
The sum of the first n terms:
\(\displaystyle S_n = \frac{n}{2}(u_1 + u_n)\).
Thus for \(n=10\):
\(\displaystyle S_{10} = \frac{10}{2}(1 + 19) = 5 \times 20 = 100.\)
Question 10 [Maximum mark: 6]
Consider an arithmetic sequence: 4, 10, 16, 22, …
- (a) Find the common difference, d.
- (b) Find the 10th term in the sequence.
- (c) Find the sum of the first 10 terms in the sequence.
Solution:
(a)
Here \(u_1 = 4\) and \(u_2 = 10\). Hence
\(d = 10 - 4 = 6.\)
(b)
Using the formula \(u_n = u_1 + (n-1)d\):
\(u_{10} = 4 + (10 - 1)\times 6 = 4 + 9\times 6 = 4 + 54 = 58.\)
(c)
Recall the sum formula:
\(\displaystyle S_n = \frac{n}{2}(u_1 + u_n)\).
For \(n=10\):
\(\displaystyle S_{10} = \frac{10}{2}\,(4 + 58) = 5 \times 62 = 310.\)
Essential Formulas for Sequences & Series
Below are the essential formulas you need to master for IB Math AA SL questions on sequences and series. These formulas are frequently tested and form the foundation for solving practice problems.
Arithmetic Sequences
nth Term Formula: \(u_n = u_1 + (n-1)d\)
Where \(u_1\) = first term, \(d\) = common difference, \(n\) = term number
Common Difference: \(d = u_2 - u_1 = u_{n+1} - u_n\)
Sum of First n Terms:
\[S_n = \frac{n}{2}(u_1 + u_n) = \frac{n}{2}(2u_1 + (n-1)d)\]Finding Term Number: Given \(u_n\), solve \(n = \frac{u_n - u_1}{d} + 1\)
Geometric Sequences
nth Term Formula: \(u_n = u_1 \cdot r^{n-1}\)
Where \(u_1\) = first term, \(r\) = common ratio, \(n\) = term number
Common Ratio: \(r = \frac{u_2}{u_1} = \frac{u_{n+1}}{u_n}\)
Sum of First n Terms: For \(r \neq 1\):
\[S_n = u_1 \cdot \frac{r^n - 1}{r - 1} = u_1 \cdot \frac{1 - r^n}{1 - r}\]Sum to Infinity: For \(|r| < 1\):
\[S_\infty = \frac{u_1}{1 - r}\]Special Series Formulas
Sum of First n Natural Numbers:
\[1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}\]Sum of Squares:
\[1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}\]Sum of Cubes:
\[1^3 + 2^3 + 3^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2\]Sigma Notation
Definition: \(\displaystyle \sum_{k=1}^{n} a_k = a_1 + a_2 + a_3 + \cdots + a_n\)
Properties:
- \(\displaystyle \sum_{k=1}^{n} c \cdot a_k = c \cdot \sum_{k=1}^{n} a_k\) (Constant multiple)
- \(\displaystyle \sum_{k=1}^{n} (a_k + b_k) = \sum_{k=1}^{n} a_k + \sum_{k=1}^{n} b_k\) (Sum rule)
- \(\displaystyle \sum_{k=1}^{n} c = nc\) (Constant sum)
Applications
Compound Interest: \(A = P(1 + r)^n\) (Geometric sequence application)
Linear Growth/Decay: \(A = P + nd\) (Arithmetic sequence application)
Loan Repayments: Combines geometric series with regular payments
Depreciation: \(V = V_0(1 - r)^n\) where \(r\) = depreciation rate
Frequently Asked Questions (FAQ)
Geometric: \(u_n = u_1 \times r^{n-1}\), \(S_n = u_1 \times \frac{r^n - 1}{r - 1}\), \(S_\infty = \frac{u_1}{1-r}\) for \(|r| < 1\)
These formulas are given in the IB formula booklet, but knowing them by heart speeds up problem-solving.
Arithmetic: Solve \(u_1 + (n-1)d = \text{value}\) for \(n\).
Geometric: Solve \(u_1 \times r^{n-1} = \text{value}\), then use logarithms: \(n = \frac{\log(\text{value}/u_1)}{\log(r)} + 1\).
Geometric: Compound interest, population growth, radioactive decay, loan repayments, bouncing ball heights.
IB exam questions often include these applications as word problems.
Arithmetic: \(u_n = u_{n-1} + d\) with \(u_1\) given.
Geometric: \(u_n = r \times u_{n-1}\) with \(u_1\) given.
This is different from the explicit formula which gives \(u_n\) directly.