IB MYP 4-5 Maths- Arithmetic and geometric sequences - Study Notes - New Syllabus
IB MYP 4-5 Maths- Arithmetic and geometric sequences – Study Notes
Extended
- Arithmetic and geometric sequences
IB MYP 4-5 Maths- Arithmetic and geometric sequences – Study Notes – All topics
Arithmetic and Geometric sequences
Arithmetic Sequences
An arithmetic sequence (or arithmetic progression) is a number pattern in which each term is obtained by adding a fixed number to the previous term. This fixed number is called the common difference.
General Form of an Arithmetic Sequence:
\( a,\ a + d,\ a + 2d,\ a + 3d,\ \ldots \)
Where:
- \( a \) = first term
- \( d \) = common difference
- \( n \) = position (term number)
Formula for the $n^{th}$ Term
The general formula to find the \(n^{th}\) term of an arithmetic sequence is:
\( T_n = a + (n – 1)d \)
Formula for the Sum of First $n$ Terms
The sum of the first \(n\) terms of an arithmetic sequence is given by:
\( S_n = \frac{n}{2} \left[2a + (n – 1)d\right] \)
or
\( S_n = \frac{n}{2}(a + l) \)
Where:
- \( l \) is the last term (can be calculated using \( l = a + (n – 1)d \))
Important Properties of Arithmetic Sequences
- The difference between any two consecutive terms is constant.
- The graph of an arithmetic sequence (with \(n\) on x-axis, term on y-axis) is a straight line.
- If the common difference is:
- Positive → sequence increases.
- Negative → sequence decreases.
- Zero → sequence is constant.
Example
The first term of an arithmetic sequence is 7 and the common difference is 3. Find the 15th term.
▶️ Answer/Explanation
Use the formula: \( T_n = a + (n – 1)d \)
\( T_{15} = 7 + (15 – 1) \cdot 3 = 7 + 14 \cdot 3 = 7 + 42 = \boxed{49} \)
Example
An arithmetic sequence has first term 4, common difference 5, and the last term is 74. Find the number of terms.
▶️ Answer/Explanation
Use the formula: \( T_n = a + (n – 1)d \)
\( 74 = 4 + (n – 1) \cdot 5 \)
\( 70 = (n – 1) \cdot 5 \Rightarrow n – 1 = 14 \Rightarrow n = \boxed{15} \)
Example
Find the sum of the first 20 terms of the sequence: 2, 5, 8, 11, …
▶️ Answer/Explanation
First term: \( a = 2 \), common difference: \( d = 3 \), number of terms: \( n = 20 \)
Use the sum formula: \( S_n = \frac{n}{2} [2a + (n – 1)d] \)
\( S_{20} = \frac{20}{2} [2 \cdot 2 + 19 \cdot 3] = 10[4 + 57] = 10 \cdot 61 = \boxed{610} \)
Example
If the 6th term of an arithmetic sequence is 20 and the 10th term is 36, find the common difference.
▶️ Answer/Explanation
Use the formula: \( T_n = a + (n – 1)d \)
\( T_{10} – T_6 = [a + 9d] – [a + 5d] = 36 – 20 = 4d \)
\( \Rightarrow d = \frac{16}{4} = \boxed{4} \)
Example
In a staircase, each step is 15 cm taller than the previous one. The height of the first step is 20 cm. What is the height of the 10th step?
▶️ Answer/Explanation
This is an arithmetic sequence with \( a = 20 \), \( d = 15 \), and \( n = 10 \)
\( T_{10} = a + (n – 1)d = 20 + 9 \cdot 15 = 20 + 135 = \boxed{155\ \text{cm}} \)
Geometric Sequences
A geometric sequence is a sequence where each term is obtained by multiplying the previous term by a fixed number called the common ratio.
This is different from an arithmetic sequence, where a fixed number is added.
General Form of a Geometric Sequence:
\( a,\ ar,\ ar^2,\ ar^3,\ ar^4,\ \ldots \)
Where:
- \( a \) is the first term
- \( r \) is the common ratio (found by dividing any term by the previous one)
Formula for the $n^{th}$ Term:
The \( n^{th} \) term of a geometric sequence is given by:
\( u_n = ar^{n-1} \)
To find the common ratio \( r \):
\( r = \frac{\text{any term}}{\text{previous term}} \)
Important Notes:
- If \( r > 1 \), the sequence is increasing.
- If \( 0 < r < 1 \), the sequence is decreasing.
- If \( r = 1 \), the sequence is constant.
- If \( r < 0 \), the sequence alternates in sign.
To find a missing term or the position of a term, use:
\( u_n = ar^{n-1} \Rightarrow n = 1 + \log_r \left(\frac{u_n}{a}\right) \)
Example:
The first three terms of a geometric sequence are: 3, 6, 12, …
Find the 6th term.
▶️ Answer/Explanation
Identify \( a \) and \( r \)
First term \( a = 3 \), common ratio \( r = \frac{6}{3} = 2 \)
Use the formula
\( u_6 = ar^{6-1} = 3 \cdot 2^5 = 3 \cdot 32 = \boxed{96} \)
Example:
The 1st term of a geometric sequence is 81 and the 4th term is 3. Find the common ratio and the 6th term.
▶️ Answer/Explanation
Use formula for 4th term
\( u_4 = ar^{3} = 3 \), and \( a = 81 \)
\( 81r^3 = 3 \Rightarrow r^3 = \frac{3}{81} = \frac{1}{27} \Rightarrow r = \frac{1}{3} \)
Find 6th term
\( u_6 = ar^5 = 81 \cdot \left(\frac{1}{3}\right)^5 = 81 \cdot \frac{1}{243} = \boxed{\frac{1}{3}} \)
Special Types:
- When \( r = -1 \): Alternating constant sequence
- When \( r = 0 \): All terms after the first are 0
- When \( 0 < r < 1 \): Terms approach 0
Geometric Series (Sum of Terms)
A geometric sequence is a number pattern in which each term is found by multiplying the previous term by a constant called the common ratio (\(r\)).
A geometric series is the sum of the terms of a geometric sequence.
General Form:
\( a,\ ar,\ ar^2,\ ar^3,\ \ldots \)
Formula for the Sum of First \(n\) Terms of a Geometric Series:
If \( r \neq 1 \), then:
\( S_n = a \frac{1 – r^n}{1 – r} \quad \text{(when } r < 1 \text{ or positive)} \) \( S_n = a \frac{r^n – 1}{r – 1} \quad \text{(when } r > 1 \text{ or negative)} \)
\( S_n \) = sum of first \(n\) terms, \( a \) = first term, \( r \) = common ratio, \( n \) = number of terms
Example:
Find the sum of the first 6 terms of the geometric series: \( 3 + 6 + 12 + 24 + \ldots \)
▶️ Answer/Explanation
\( a = 3 \), \( r = \frac{6}{3} = 2 \), \( n = 6 \)
Use the formula:
\( S_6 = 3 \cdot \frac{2^6 – 1}{2 – 1} = 3 \cdot \frac{64 – 1}{1} = 3 \cdot 63 = 189 \)
Example:
Find the sum of the first 4 terms of the geometric series: \( 81 + 27 + 9 + 3 + \ldots \)
▶️ Answer/Explanation
\( a = 81 \), \( r = \frac{27}{81} = \frac{1}{3} \), \( n = 4 \)
Use the formula:
\( S_4 = 81 \cdot \frac{1 – \left(\frac{1}{3}\right)^4}{1 – \frac{1}{3}} \)
\( S_4 = 81 \cdot \frac{1 – \frac{1}{81}}{\frac{2}{3}} = 81 \cdot \frac{\frac{80}{81}}{\frac{2}{3}} \)
\( S_4 = \frac{80}{1} \cdot \frac{3}{2} = 120 \)
Example:
Find the sum of the first 5 terms of the geometric series: \( 5 – 10 + 20 – 40 + \ldots \)
▶️ Answer/Explanation
\( a = 5 \), \( r = -2 \), \( n = 5 \)
Use the formula:
\( S_5 = 5 \cdot \frac{(-2)^5 – 1}{-2 – 1} = 5 \cdot \frac{-32 – 1}{-3} = 5 \cdot \frac{-33}{-3} = 5 \cdot 11 = 55 \)
Key Points to Remember
- Geometric sequences multiply by a fixed value each time (common ratio).
- To use the sum formula, identify \( a \), \( r \), and \( n \) clearly.
- If \( |r| < 1 \), the terms get smaller — common in fractional ratio series.
- If \( r \) is negative, terms alternate in sign (positive/negative).